Maths Chapter-08: Binomial Theorem

Binomial Theorem:

[a + b]ⁿ = [ ⁿC₀× aⁿ] + [ ⁿC₁× (a^{n – 1}) × b ] + [ ⁿC₂× (a^{n – 2}) × b²] + [ ⁿC₃× (a^{n – 3})× b³] + . . . . . . . . . . . . . + [ ⁿC_{n – 1}× a × (b^{n – 1}) ] + [ ⁿC_n× bⁿ]

\(\Rightarrow\) Binomial Coefficient:

The coefficients ⁿC₀, ⁿC_1,ⁿC₂ . . . . . . . . . ⁿC_n occurring in the Binomial Theorem are known as Binomial coefficients

Some conclusions from Binomial Theorem:

(i) [x + y]ⁿ = [ ⁿC₀× (xⁿ) ] + [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] + [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n× yⁿ]

(ii) [x – y]ⁿ = [ ⁿC₀× (xⁿ) ] – [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] – [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ[ ⁿC_n× yⁿ]

(iii) [1 – x]ⁿ = [ ⁿC₀] – [ ⁿC₁. x ] + [ ⁿC₂. x²] – [ ⁿC₃. x³] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ[ ⁿC_n. xⁿ]

(iv) (a + b)ⁿ= \(\sum_{r\;=\;0}^{n}\) ⁿC_r (a)^{n – r}× b^r

NOTE:

(1.) ⁿC_r = \(\frac{n!}{r!(n-r)!}\) where, n is a non-negative integer and [0 ≤ r ≤ n]

(2.) ⁿC₀= ⁿC_n= 1

(3.) There are total (n + 1) terms in the expansion of (a + b)ⁿ

Exercise 8.1

Q.1: Expand the Expression (1 – 3x)⁵

Sol.

By using Binomial Theorem:

Since, (1 – x)ⁿ = [ ⁿC₀ ] + [ ⁿC₁× (-x) ] + [ ⁿC₂× (-x)²] + [ ⁿC₃× (-x³) ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n× (-x)ⁿ]

Therefore, (1 – 3x)⁵ = [ ⁵C₀] – [ ⁵C₁× (3x) ] + [ ⁵C₂× (3x)²] – [ ⁵C₃× (3x)³] + [ ⁵C₄× (3x)⁴ ] – [ ⁵C₅× (3x)⁵ ]

\(\\\Rightarrow\) 1 – [ 5 × (3x) ] + [ 10 × (9x²) ] – [ 10 × (27x³) ] + [ 5 × (81x⁴) ] – [ 1 × (243x⁵) ]

\(\\\Rightarrow\) 1 – 15x + 90x² – 270x³ + 405x⁴ – 243x⁵

Therefore, (1 – 3x)⁵ = 1 – 15x + 90x² – 270x³ + 405x⁴ – 243x⁵

Q.2: Expand the Expression \(\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}\)

Sol.

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀× (xⁿ) ] – [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] – [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ[ ⁿC_n× yⁿ]

Therefore, \(\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}:\)

[⁵C₀× \(\left ( \frac{5}{x} \right )^{5}\) ] – [⁵C₁× \(\left ( \frac{5}{x} \right )^{4}\) × \(\left ( \frac{x}{5} \right )^{1}\) ] + [⁵C₂× \(\left ( \frac{5}{x} \right )^{3}\) × \(\left ( \frac{x}{5} \right )^{2}\) ] – [ ⁵C₃× \(\left ( \frac{5}{x} \right )^{2}\) × \(\left ( \frac{x}{5} \right )^{3}\) ] + [⁵C₄× \(\left ( \frac{5}{x} \right )^{1}\) × \(\left ( \frac{x}{5} \right )^{4}\) ] – [⁵C₅× \(\left ( \frac{x}{5} \right )^{5}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{3125}{x^{5}}\right ]-\left [ 5\times \frac{625}{x^{4}}\times \frac{x}{5}\right ]+\left [ 10\times \frac{125}{x^{3}}\times \frac{x^{2}}{25}\right ]-\left [ 10\times \frac{25}{x^{2}}\times \frac{x^{3}}{125} \right ]+\left [ 5\times \frac{5}{x}\times \frac{x^{4}}{625} \right ]-\left [ 1\times \frac{x^{5}}{3125} \right ]\\\) \(\\\Rightarrow \left [\frac{3125}{x^{5}}\right ]-\left [\frac{625}{x^{3}}\right ]+\left [ \frac{50}{x}\right ]-2x+\left [ \frac{3x^{3}}{25} \right ]-\left [\frac{x^{5}}{3125} \right ]\\\)

Therefore, \(\\\left ( \frac{5}{x}-\frac{x}{5} \right )^{5}\):

\(\\\left [\frac{x^{5}}{3125} \right ] +\left [ \frac{3x^{3}}{25} \right ]-2x+\left [ \frac{50}{x}\right ]-\left [\frac{625}{x^{3}}\right ]+ \left [\frac{3125}{x^{5}}\right ]\)

Q.3: Expand the Expression (3x – 2)⁶

Sol.

By using Binomial Theorem:

[x – y]ⁿ = [ ⁿC₀× (xⁿ) ] – [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] – [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . . . . . . +(-1)ⁿ[ ⁿC_n× yⁿ]

Therefore, (3x – 2)⁶= [ ⁶C₀× (3x)⁶] – [ ⁶C₁× (3x)⁵× (2) ] + [ ⁶C₂× (3x)⁴× (2)²] – [ ⁶C₃× (3x)³× (2)³] + [ ⁶C₄× (3x)²× (2)⁴] – [ ⁶C₅× (3x)¹× (2)⁵] + [ ⁶C₆ × (2)⁶]

\(\\\Rightarrow\) [1 × (729x⁶)] – [6 × (243x⁵) × 2] + [15 × (81x⁴) × 4] – [20 × (27x³) × 8 ] + [15 × (9x²) × 16] – [6 × (3x) × 32] + [1 × 64]

\(\\\Rightarrow\) 729x⁶ – 2916x⁵ + 4860x⁴ – 4320x³ + 2160x² – 576x + 64

Therefore, (3x – 2)⁶= 729x⁶ – 2916x⁵ + 4860x⁴ – 4320x³ + 2160x² – 576x + 64

Q.4: Expand the Expression \(\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}\)

Sol.

By using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀× (xⁿ) ] + [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] – [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n× yⁿ]

Therefore, \(\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}\):

[ ⁵C₀× \(\left ( \frac{4}{x} \right )^{5}\) ] + [ ⁵C₁× \(\left ( \frac{4}{x} \right )^{4}\) × \(\left ( \frac{x}{3} \right )^{1}\) ] + [ ⁵C₂× \(\left ( \frac{4}{x} \right )^{3}\) × \(\left ( \frac{x}{3} \right )^{2}\) ] + [ ⁵C₃× \(\left ( \frac{4}{x} \right )^{2}\) × \(\left ( \frac{x}{3} \right )^{3}\) ] + [ ⁵C₄× \(\left ( \frac{4}{x} \right )^{1}\) × \(\left ( \frac{x}{3} \right )^{4}\) ] + [ ⁵C₅× \(\left ( \frac{x}{3} \right )^{5}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{1024}{x^{5}}\;\right ]+\left [ 5\times \frac{256}{x^{4}}\times \frac{x}{3}\;\right ]+\left [ 10\times \frac{64}{x^{3}}\times \frac{x^{2}}{9}\;\right ]+\left [ 10\times \frac{16}{x^{2}}\times \frac{x^{3}}{27} \right ]+\left [ 5\times \frac{4}{x}\times \frac{x^{4}}{81}\; \right ]+\left [ 1\times \frac{x^{5}}{243} \right ]\\\\\) \(\\\Rightarrow \left [\frac{1024}{x^{5}}\right ]+\left [\frac{1280}{3x^{3}}\right ]+\left [ \frac{640}{9x}\right ]+\left [\frac{160x}{27}\right ]+\left [ \frac{20x^{3}}{81} \right ]+\left [\frac{x^{5}}{243} \right ]\\\)

Therefore, \(\left ( \frac{4}{x}+\frac{x}{3} \right )^{5}:\)

\(\\\Rightarrow \left [\frac{x^{5}}{243} \right ]+\left [ \frac{20x^{3}}{81} \right ]+\left [\frac{160x}{27}\right ]+\left [ \frac{640}{9x}\right ]+\left [\frac{1280}{3x^{3}}\right ]+\left [\frac{1024}{x^{5}}\right ]\)

Q.5: Expand the Expression \(\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}\)

Sol.

By using Binomial Theorem:

Therefore, \(\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}\):

[ ⁶C₀× \(\left ( \frac{1}{x} \right )^{6}\) ] – [ ⁶C₁× \(\left ( \frac{1}{x} \right )^{5}\) × \(\left ( \frac{x}{2} \right )^{1}\) ] + [ ⁶C₂× \(\left ( \frac{1}{x} \right )^{4}\) × \(\left ( \frac{x}{2} \right )^{2}\) ] – [ ⁶C₃× \(\left ( \frac{1}{x} \right )^{3}\) × \(\left ( \frac{x}{2} \right )^{3}\) ] + [ ⁶C₄× \(\left ( \frac{1}{x} \right )^{2}\) × \(\left ( \frac{x}{2} \right )^{4}\) ] – [ ⁶C₅× \(\left ( \frac{1}{x} \right )^{1}\) × \(\left ( \frac{x}{2} \right )^{5}\) ] + [ ⁶C₆×\(\left ( \frac{x}{2} \right )^{6}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{1}{x^{6}}\;\right ]-\left [ 6\times \frac{1}{x^{5}}\times \frac{x}{2}\;\right ]+\left [ 15\times \frac{1}{x^{4}}\times \frac{x^{2}}{4}\;\right ]-\left [ 20\times \frac{1}{x^{3}}\times \frac{x^{3}}{8} \right ]+\left [ 15\times \frac{1}{x^{2}}\times \frac{x^{4}}{16}\; \right ]-\left [ 6\times \frac{1}{x}\times \frac{x^{5}}{32} \right ]+\left [ 1\times \frac{x^{6}}{64} \right ]\\\) \(\\\Rightarrow \left [ \frac{1}{x^{6}}\;\right ]-\left [ \frac{3}{x^{4}}\right ]+\left [ \frac{15}{4x^{2}}\right ]-\left [ \frac{5}{2} \right ]+\left [\frac{15x^{2}}{16}\; \right ]-\left [\frac{3x^{4}}{16} \right ]+\left [\frac{x^{6}}{64} \right ]\\\)

Therefore, \(\left ( \frac{1}{x}-\frac{x}{2} \right )^{6}:\)

\(\\\Rightarrow \left [\frac{x^{6}}{64} \right ]-\left [\frac{3x^{4}}{16} \right ]+\left [\frac{15x^{2}}{16} \right ]-\left [ \frac{5}{2} \right ]+\left [ \frac{15}{4x^{2}}\right ]-\left [ \frac{3}{x^{4}}\right ]+\left [ \frac{1}{x^{6}}\right ]\)

Q.6: By using Binomial Theorem, Evaluate (98)⁴

Sol.

(98)⁴ = (100 – 2)⁴

Now, by using Binomial Theorem:

Therefore, (100 – 2)⁴ = [ ⁴C₀× (100)⁴] – [ ⁴C₁× (100)³ × (2) ] + [ ⁴C₂× (100)² × (2)²] – [ ⁴C₃ × 100 × (2)³] + [⁴C₄× (2)⁴]

\(\\\Rightarrow\) (1 × 100000000) – (4 × 1000000 × 2) + (6 × 10000 × 4) – (4 × 100 × 8 ) + (1 × 16)

\(\\\Rightarrow\) 100000000 – 8000000 + 240000 – 3200 + 16

Therefore, (98)⁴ = (100 – 2)⁴= 92236816 = 9.2236816 × 10⁷

Q.7: By using Binomial Theorem, Evaluate (105)⁵

Sol.

(105)⁵ = (100 + 5)⁵

Now, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀× (xⁿ) ] + [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] + [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n× yⁿ]

Therefore, (100 + 5)⁵ = [ ⁵C₀× (100)⁵] + [ ⁵C₁× (100)⁴ × (5) ] + [ ⁵C₂× (100)³ × (5)²] + [ ⁵C₃ × (100)² × (5)³] + [ ⁵C₄× (100) × (5)⁴ ] + [ ⁵C₅ × (5)⁵]

\(\\\Rightarrow\) (1 × 10000000000) + (5 × 100000000 × 5) + (10 × 1000000 × 5²) + (10 × 10000 × 5³ ) + (5 × 100 × 5⁴ ) + ( 1 × 5⁵ )

\(\\\Rightarrow\) 10000000000 + 2500000000 + 250000000 + 12500000 + 312500 + 3125

Therefore, (105)⁵ = (100 + 5)⁵ = 12762815630 = 1.276281563 × 10¹⁰

Q.8: By using Binomial Theorem, Evaluate (104)⁴

Sol.

(104)⁴ = (100 + 4)⁴

Now, by using Binomial Theorem:

Therefore, (100 + 4)⁴ = [⁴C₀× (100)⁴] + [ ⁴C₁× (100)³ × (4) ] + [⁴C₂× (100)² × (4)²] + [⁴C₃ × 100 × (4)³] + [⁴C₄ × (4)⁴]

\(\\\Rightarrow\) (1 × 100000000) + (4 × 1000000 × 4) + (6 × 10000 × 16) + (4 × 100 × 64 ) + (1 × 256)

\(\\\Rightarrow\) 100000000 + 16000000 + 960000 + 25600 + 256

Therefore, (104)⁴ = (100 + 4)⁴= 116985856 = 1.16985856 × 10⁸

Q.9: By using Binomial Theorem, Evaluate (95)⁵

Sol.

(95)⁵ = (100 – 5)⁵

Now, by using Binomial Theorem:

Therefore, (100 – 5)⁵ = [ ⁵C₀× (100)⁵] – [ ⁵C₁× (100)⁴ × (5) ] + [ ⁵C₂× (100)³ × (5)²] – [ ⁵C₃ × (100)² × (5)³] + [ ⁵C₄× (100) × (5)⁴ ] – [ ⁵C₅ × (5)⁵ ]

\(\\\Rightarrow\) (1 × 10000000000) – (5 × 100000000 × 5) + (10 × 1000000 × 5²) – (10 × 10000 × 5³ ) + (5 × 100 × 5⁴ ) – ( 1 × 5⁵ )

\(\\\Rightarrow\) 10000000000 – 2500000000 + 250000000 – 12500000 + 312500 – 3125

Therefore, (95)⁵ = (100 – 5)⁵ = 7737809375 = 7.737809375 × 10⁹

Q.10: By using Binomial Theorem, determine which number is greater (1.1)¹⁰⁰⁰⁰ or 1000.

Sol.

Now, (1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰

Now, on applying Binomial Theorem, the first few terms in the expansion of (1 + 0.1)¹⁰⁰⁰⁰ are:

Since, [x + y]ⁿ = [ ⁿC₀× (xⁿ) ] + [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] – [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n× yⁿ]

Therefore, (1 + 0.1)¹⁰⁰⁰⁰= [ ¹⁰⁰⁰⁰C₀× (1)¹⁰⁰⁰⁰ ] + [ ¹⁰⁰⁰⁰C₁× (1)⁹⁹⁹⁹ × 0.1 ] + other positive terms

\(\\\Rightarrow\) [1 × 1] + [10000 × 1 × 0.1] + other positive terms

\(\\\Rightarrow\) 1 + 1000 + other positive terms > 1000

Therefore, (1.1)¹⁰⁰⁰⁰ is greater than 1000.

Q.11: Find (a + b)⁵ – (a – b)⁵. Hence evaluate \(\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5}\)

Sol.

Now, by using Binomial Theorem:

And, [x – y]ⁿ = [ ⁿC₀× (xⁿ) ] – [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] – [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . . . . . . + (-1)ⁿ[ ⁿC_n× yⁿ]

Therefore, (a + b)⁵= [ ⁵C₀× (a⁵) ] + [ ⁵C₁× (a⁴) × b ] + [ ⁵C₂× (a³) × b²] + [ ⁵C₃× (a²) × b³] + [ ⁵C₄× (a¹) × b⁴] + [ ⁵C₅× b⁵]

And, [a – b]⁵ = [ ⁵C₀× (a⁵) ] – [ ⁵C₁× (a⁴) × b ] + [ ⁵C₂× (a³) × b²] – [ ⁵C₃× (a²) × b³] + [ ⁵C₄× (a¹) × b⁴] – [ ⁵C₅× b⁵]

Therefore, (a + b)⁵– (a – b)⁵ = [ ⁵C₀a⁵ + ⁵C₁ a⁴b + ⁵C₂a³b² + ⁵C₃a²b³+ ⁵C₄ab⁴ + ⁵C₅b⁵] – [ ⁵C₀a⁵ – ⁵C₁a⁴b + ⁵C₂a³b² – ⁵C₃a²b³+ ⁵C₄ab⁴ – ⁵C₅b⁵]

\(\\\Rightarrow\) [ ⁵C₀a⁵ + ⁵C₁ a⁴b + ⁵C₂a³b² + ⁵C₃a²b³+ ⁵C₄ab⁴ + ⁵C₅b⁵ – ⁵C₀a⁵ + ⁵C₁a⁴b – ⁵C₂a³b² + ⁵C₃a²b³– ⁵C₄ab⁴ + ⁵C₅b⁵]

\(\\\Rightarrow\) 2[ ⁵C₁a⁴b + ⁵C₃a²b³+⁵C₅b⁵] = 2[ 5a⁴b + 10a²b³+ b⁵ ]

Therefore, (a + b)⁵– (a – b)⁵ = 2b[ 5a⁴+ 10a²b² + b⁴] . . . . . . (1)

Now, on substituting a = \(\sqrt{5}\) and b = \(\sqrt{3}\) in equation (1) we will get:

\(\\\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5} = 2\sqrt{3}\times \left [ 5( \sqrt{5})^{4} +10( \sqrt{5})^{2}\times ( \sqrt{3} )^{2}+(\sqrt{3})^{4}\right ]\\\)

\(\\\Rightarrow 2\sqrt{3}\;[125+150+9]=568\sqrt{3}\\\)

Therefore, \(\left(\sqrt{5}+\sqrt{3} \right )^{5}-\left ( \sqrt{5}-\sqrt{3}\right)^{5}\)=\(568\sqrt{3}\)

Q.12: Find (1 + x)⁵ – (1 – x)⁵. Hence evaluate \(\left ( 1+\sqrt{7} \right )^{5}-\left ( 1-\sqrt{7} \right )^{5}\)

Sol.

Now, by using Binomial Theorem:

Since, [x + y]ⁿ = [ ⁿC₀× (xⁿ) ] + [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] + [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . + [ ⁿC_n× yⁿ]

And, [x – y]ⁿ = [ ⁿC₀× (xⁿ) ] – [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] – [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . + (-1)ⁿ[ ⁿC_n× yⁿ]

Therefore, (1 + x)⁵= [ ⁵C₀× (1⁵) ] + [ ⁵C₁× (1⁴) × (x) ] + [ ⁵C₂× (1³) × (x)²] + [ ⁵C₃× (1²) × (x)³] + [ ⁵C₄× (1¹) × (x)⁴] + [ ⁵C₅× (x)⁵]

And, [1 – x]⁵ = [ ⁵C₀× (1⁵) ] – [ ⁵C₁× (1⁴) × (x) ] + [ ⁵C₂× (1³) × (x)²] – [ ⁵C₃× (1²) × (x)³] + [ ⁵C₄× (1¹) × (x)⁴] – [ ⁵C₅× (x)⁵]

Therefore, (1 + x)⁵– (1 – x)⁵ = [ ⁵C₀ + ⁵C₁ x + ⁵C₂x² + ⁵C₃x³+ ⁵C₄x⁴ + ⁵C₅x⁵] – [ ⁵C₀ – ⁵C₁x + ⁵C₂x² – ⁵C₃x³+ ⁵C₄x⁴ – ⁵C₅x⁵]

\(\\\Rightarrow\) [ ⁵C₀ + ⁵C₁ x + ⁵C₂x² + ⁵C₃x³+ ⁵C₄x⁴ + ⁵C₅x⁵ – ⁵C₀ + ⁵C₁x – ⁵C₂x² + ⁵C₃x³– ⁵C₄x⁴ + ⁵C₅x⁵]

\(\\\Rightarrow\) 2[ ⁵C₁x + ⁵C₃x³ + ⁵C₅x⁵] = 2[ 5x + 10x³+ x⁵ ]

Therefore, (1 + x)⁵– (1 – x)⁵ = 2x[ 5 + 10x² + x⁴] . . . . . . (1)

Now, on substituting x = \(\sqrt{7}\) in equation (1) we will get:

\(\\\Rightarrow \left ( 1+\sqrt{7} \right )^{7}-\left ( 1-\sqrt{7} \right )^{5}=2\sqrt{7}\left [ 5+10\left ( \sqrt{7} \right )^{2} + \left ( \sqrt{7} \right )^{4}\right ]\\\) \(\\\Rightarrow 2\sqrt{7}\;[5+70+49]=248\sqrt{7}\\\)

Therefore, \(\left(1+\sqrt{7} \right )^{5}-\left ( 1-\sqrt{7}\right)^{5}\)=\(\;248\sqrt{7}\)

Q.13 Show that 7ⁿ⁺¹ – 6n – 7 is divisible by 36, where n is the positive integer.

Sol.

Equation 7ⁿ⁺¹ – 6n – 7 will be divisible by 36 if, 7ⁿ⁺¹ – 6n – 7 = 36p [where p is any natural number]

Now, by Binomial Theorem:

Since, [1 + x]^m = [ ^mC₀] + [ ^mC₁. x ] + [ ^mC₂. x²] + [ ^mC₃. x³] + . . . . . . . . . . . . + [ ^mC_m. x^m]

Therefore, for x = 6 and m = n + 1 the equation becomes:

[1 + 6]ⁿ⁺¹ = [ ⁿ⁺¹C₀] + [ ⁿ⁺¹C₁× 6 ] + [ ⁿ⁺¹C₂× (6)²] + [ ⁿ⁺¹C₃× (6)³] + . . . . . . . . . . . . . + [ ⁿ⁺¹C_n+1× (6)ⁿ⁺¹]

\(\\\Rightarrow\) [7]^{n + 1} = 1 + [(n + 1) × 6] + 6²[ ⁿ⁺¹C₂+ ⁿ⁺¹C₃ × (6) + . . . . . . . . . +ⁿ⁺¹C_{n + 1}× (6)^n–1]

\(\\\Rightarrow\) [7]^{n + 1} = 1 + 6n + 6 + 36 [ ⁿ⁺¹C₂+ ⁿ⁺¹C₃ × (6) + . . . . . . . . . +^{n + 1}C_{n + 1}× (6)^{n – 1}]

\(\\\Rightarrow\) (7)^{n + 1} – 6 n – 7 = 36p

Where p is any natural number and p = [^{n + 1}C₂+ ^{n + 1}C₃ × (6) + . . . . . . . . . . . . . . . . . . +^{n + 1}C_{n + 1}× (6)^{n – 1}]

Therefore, 7ⁿ⁺¹ – 6n – 7 is divisible by 36 [where n is a positive integer]

Q.14 Prove that \(\sum_{r\;=\;0}^{n}\) 2^r × ⁿC_r = 3ⁿ

Sol.

By using Binomial Theorem:

Since, (a + b)ⁿ= \(\sum_{r\;=\;0}^{n}\) ⁿC_r (a)^{n – r}× (b)^r

Therefore, on substituting the value of a = 1 and b = 2 in the above equation we will get:

(1 + 2)ⁿ= \(\sum_{r\;=\;0}^{n}\) ⁿC_r (1)^{n – r}× (2)^r

\(\\\Rightarrow\) 3ⁿ= \(\sum_{r\;=\;0}^{n}\) ⁿC_r × (2)^r

Therefore, \(\\\sum_{r\;=\;0}^{n}\) 2^r × ⁿC_r = 3ⁿ

Q.15 Show that 5ⁿ – 4n always leaves remainder 5 when divided by 16, where n is the positive integer.

Sol.

Equation 5ⁿ – 4n will leave remainder 5 when divided by 16 if, 5ⁿ – 4n = 16p + 5 [where p is any natural number]

Now, by Binomial Theorem:

Since, [1 + x]ⁿ = [ ⁿC₀] + [ ⁿC₁. x ] + [ ⁿC₂. x²] + [ ⁿC₃. x³] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n. xⁿ]

Therefore, for x = 4 the equation becomes:

[1 + 4]ⁿ = [ ⁿC₀] + [ ⁿC₁× 4 ] + [ ⁿC₂× (4)²] + [ ⁿC₃× (4)³] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n× (4)ⁿ]

\(\\\Rightarrow\) [5]ⁿ = 1 + [(n + 1) × 4] + 4²[ ⁿC₂+ ⁿC₃ × (4) + . . . . . . . . . . . . . . . . . . +ⁿC_n× (4)ⁿ]

\(\\\Rightarrow\) [5]ⁿ = 1 + 4n + 4 + 16 [ ⁿC₂+ ⁿC₃ × (4) + . . . . . . . . . . . . . . . . . . +ⁿC_n× (4)^n–1]

\(\\\Rightarrow\) 5ⁿ – 4n – 5 = 16p

Where p is any natural number and p = [ ⁿC₂+ ⁿC₃ × (4) + . . . . . . . . . . . . . . . . . . +ⁿC_n× (4)^n–1]

i.e. 5ⁿ – 4n = 16 + 5

Therefore, 5ⁿ – 4n will leave remainder 5 when divided by 16, where n is any natural number.

Q.16: Find (a + b)⁶ – (a – b)⁶. Hence evaluate: \(\left(\sqrt{2}+\sqrt{3} \right )^{6}-\left ( \sqrt{2}-\sqrt{3}\right)^{6}\)

Sol.

Now, by using Binomial Theorem:

Therefore, [a + b]⁶= [ ⁶C₀× (a⁶) ] + [ ⁶C₁× (a⁵) × b ] + [ ⁶C₂× (a⁴) × b²] + [ ⁶C₃× (a³) × b³] + [ ⁶C₄× (a²) × b⁴] + [ ⁶C₅× a × b⁵] + [⁶C₆ × b⁶]

And, [a – b]⁶ = [ ⁶C₀× (a⁶) ] – [ ⁶C₁× (a⁵) × b ] + [ ⁶C₂× (a⁴) × b²] – [ ⁶C₃× (a³) × b³] + [ ⁶C₄× (a²) × b⁴] – [ ⁶C₅× a × b⁵] + [⁶C₆ × b⁶]

Therefore, (a + b)⁶– (a – b)⁶ = [ ⁶C₀ a⁶ + ⁶C₁a⁵b + ⁶C₂a⁴b²+ ⁶C₃a³b³ + ⁶C₄a²b⁴ + ⁶C₅ab⁵ + ⁶C₆ b⁶] – [ ⁶C₀a⁶ – ⁶C₁a⁵b + ⁶C₂a⁴b² – ⁶C₃ a³b³+ ⁶C₄a²b⁴ – ⁶C₅ab⁵ + ⁶C₆ b⁶]

\(\\\Rightarrow\) [ ⁶C₀ a⁶ + ⁶C₁a⁵b + ⁶C₂a⁴b²+ ⁶C₃a³b³ + ⁶C₄a²b⁴ + ⁶C₅ab⁵ + ⁶C₆ b⁶ – ⁶C₀a⁶ + ⁶C₁a⁵b – ⁶C₂a⁴b² + ⁶C₃ a³b³– ⁶C₄a²b⁴ + ⁶C₅ab⁵ – ⁶C₆ b⁶]

\(\\\Rightarrow\) 2[ ⁶C₁a⁵b + ⁶C₃a³b³+⁶C₅ab⁵] = 2[ 6a⁵b + 20a³b³+ 6ab⁵ ]

Therefore, (a + b)⁶– (a – b)⁶ = 4ab[ 3a⁴ + 10a²b² + 3b⁴] . . . (1)

Now, on substituting a = \(\sqrt{2}\) and b = \(\sqrt{3}\) in equation (1) we will get:

\(\\(\sqrt{2}+\sqrt{3} )^{6}-(\sqrt{2}-\sqrt{3})^{6}=4 (\sqrt{2}. \sqrt{3})\left [ 3( \sqrt{2})^{4} +10 ( \sqrt{2})^{2}\times( \sqrt{3} )^{2}+3(\sqrt{3})^{4} \right ]\\\) \(\\\Rightarrow 4\sqrt{6}\;[12+60+27]=396\sqrt{6}\\\)

Therefore, \(\left(\sqrt{2}+\sqrt{3} \right )^{6}-\left ( \sqrt{2}-\sqrt{3}\right)^{6}\)=\(\;396\sqrt{6}\)

Q.17: By using Binomial Theorem, Evaluate (91)⁴

Sol.

(91)⁴ = (100 – 9)⁴

Now, by using Binomial Theorem:

Therefore, (100 – 9)⁴ = [ ⁴C₀× (100)⁴] – [ ⁴C₁× (100)³ × (9) ] + [ ⁴C₂× (100)² × (9)²] – [ ⁴C₃ × 100 × (9)³] + [⁴C₄× (9)⁴]

\(\\\Rightarrow\) (1 × 100000000) – (4 × 1000000 × 9) + (6 × 10000 × 81) – (4 × 100 × 729) + (1 × 6561)

\(\\\Rightarrow\) 100000000 – 36000000 + 4860000 – 291600 + 6561

Therefore, (91)⁴ = (100 – 9)⁴= 68574961 = 6.8574961 × 10⁷

Q.18: By using Binomial Theorem, Evaluate (107)⁵

Sol.

(107)⁵ = (100 + 7)⁵

Now, by using Binomial Theorem:

Therefore, (100 + 7)⁵ = [ ⁵C₀× (100)⁵] + [ ⁵C₁× (100)⁴ × (7) ] + [ ⁵C₂× (100)³ × (7)²] + [ ⁵C₃ × (100)² × (7)³] + [ ⁵C₄× (100) × (7)⁴ ] + [ ⁵C₅ × (7)⁵]

\(\\\Rightarrow\) (1 × 10000000000) + (5 × 100000000 × 7) + (10 × 1000000 × 7²) + (10 × 10000 × 7³ ) + (5 × 100 × 7⁴ ) + ( 1 × 7⁵ )

\(\\\Rightarrow\) 10000000000 + 3500000000 + 490000000 + 34300000 + 1200500 + 16807

Therefore, (107)⁵ = (100 + 7)⁵ = 1402551731 = 1.402551731 × 10¹⁰

Q.19: Expand the Expression \(\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}\)

Sol.

By using Binomial Theorem:

Therefore, \(\\\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}:\\\)

[⁶C₀× \(\left ( \frac{3}{2x} \right )^{6}\) ] – [⁶C₁× \(\left ( \frac{3}{2x} \right )^{5}\) × \(\left ( \frac{x}{5} \right )^{1}\) ] + [⁶C₂× \(\left ( \frac{3}{2x} \right )^{4}\) × \(\left ( \frac{x}{5} \right )^{2}\) ] – [⁶C₃× \(\left ( \frac{3}{2x} \right )^{3}\) × \(\left ( \frac{x}{5} \right )^{3}\) ] + [⁶C₄× \(\left ( \frac{3}{2x} \right )^{2}\) × \(\left ( \frac{x}{5} \right )^{4}\) ] – [⁶C₅× \(\left ( \frac{3}{2x} \right )^{1}\) × \(\left ( \frac{x}{5} \right )^{5}\) ] + [⁶C₆×\(\left ( \frac{x}{5} \right )^{6}]\\\)

\(\\\Rightarrow \left [ 1\times \frac{729}{64\times x^{6}}\;\right ]-\left [ 6\times \frac{243}{32\times x^{5}}\times \frac{x}{5}\;\right ]+\left [ 15\times \frac{81}{16\times x^{4}}\times \frac{x^{2}}{25}\;\right ]-\left [ 20\times \frac{27}{8x^{3}}\times \frac{x^{3}}{125} \right ]+\left [ 15\times \frac{9}{4x^{2}}\times \frac{x^{4}}{625}\; \right ]-\left [ 6\times \frac{3}{2x}\times \frac{x^{5}}{3125} \right ]+\left [ 1\times \frac{x^{6}}{15625} \right ]\\\) \(\\\Rightarrow \left [\frac{729}{64\; x^{6}}\;\right ]-\left [\frac{729}{80\;x^{4}}\;\right ]+\left [\frac{243}{80\;x^{2}}\;\right ]-\left [\frac{27}{50}\right ]+\left [\frac{27x^{2}}{500}\;\right ]-\left [\frac{9x^{4}}{3125} \right ]+\left [\frac{x^{6}}{15625} \right ]\\\)

Therefore, \(\left ( \frac{3}{2x}-\frac{x}{5} \right )^{6}:\\\)

\(\Rightarrow \left [\frac{x^{6}}{15625} \right ]-\left [\frac{9x^{4}}{3125} \right ]+\left [\frac{27x^{2}}{500}\;\right ]-\left [\frac{27}{50}\right ]+\left [\frac{243}{80\;x^{2}}\;\right ]-\left [\frac{729}{80\;x^{4}}\;\right ]+\left [\frac{729}{64\; x^{6}}\;\right ]\)

Q.20: Expand the Expression (5x – 3)⁶

Sol.

By using Binomial Theorem:

Therefore, (5x – 3)⁶= [ ⁶C₀× (5x)⁶] – [ ⁶C₁× (5x)⁵× (3) ] + [ ⁶C₂× (5x)⁴× (3)²] – [ ⁶C₃× (5x)³× (3)³] + [ ⁶C₄× (5x)²× (3)⁴] – [ ⁶C₅× (5x)¹× (3)⁵] + [ ⁶C₆ × (3)⁶]

\(\\\Rightarrow\) [1 × (15625 x⁶)] – [6 × (3125 x⁵) × 3] + [15 × (625 x⁴) × 9] – [20 × (125 x³) × 27 ] + [15 × (25 x²) × 81] – [6 × (5x) × 243] + [1 × 729]

\(\\\Rightarrow\) 15625 x⁶ – 56250 x⁵ + 84375 x⁴ – 67500 x³ + 30375 x² – 7290 x + 729

Therefore, (5x – 3)⁶= 15625 x⁶ – 56250 x⁵ + 84375 x⁴ – 67500 x³ + 30375 x² – 7290 x + 729

Exercise 8.2

Formulas:

1. The general term in the expansion of (a + b)ⁿ:

T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

2. The middle term in the expansion of (a + b)ⁿ :

(a). If n is even:

The middle term = \(\left ( \frac{n}{2}+1 \right )^{th}term\)

(b). If n is odd:

The middle term = \(\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term\)

Q.1: Find the Coefficient of x⁶in the expansion of (x + 2)⁹

Sol.

Since, The general term in the expansion of (a + b)ⁿis given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

Therefore, on substituting n = 9, a = x and b = 2 in the above expression we will get:

T_{r + 1} = ⁹C_r× (x)^{9 – r} × 2^r . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x⁶, we will get:

i.e. (x)^{9 – r} = x⁶

(or) 9 – r = 6

Therefore, r = 3

Now, on substituting the value of r in equation (1) we will get:

T₄ = ⁹C₃× (x)^{9 – 3} × 2³

\(\\\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times x^{6}\times 8\\\\\) \(\\\\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 8\;x^{6}= 672\;x^{6}\\\\\)

Therefore, the Coefficient of x⁶in the expansion of (x + 2)⁹ = 672

Q.2: Find the Coefficient of a⁶b⁸in the expansion of (2a – 3b)¹⁴

Sol.

Since, The general term in the expansion of (x + y)ⁿis given by: T_{r + 1} = ⁿC_r× (x)^{n – r} × y^r

Therefore, on substituting n = 14, x = 2a and y = (-3b) in the above expression we will get:

T_{r + 1} = ¹⁴C_r× (2a)^{14 – r} × (-3b)^r

i.e. T_{r + 1} = [¹⁴C_r× (2)^{14 – r} (-3)^r ] (a)^{14 – r} × (b)^r . . . . . . . . . . (1)

Now, on comparing the coefficients of a and b in equation (1) with a⁶b⁸, we will get:

i.e. a⁶ b⁸ = a^{14 – r} b^r

Therefore, r = 8

Now, on substituting the value of ‘r’ in equation (1) we will get:

T₉ = [ ¹⁴C₈× (2)^{14 – 8} (-3)⁸ ] (a)^{14 – 8} × (b)⁸

i.e. T₉ = [ ¹⁴C₈× (2)⁶ (-3)⁸ ] a⁶ b⁸

\(\\\\\Rightarrow T_{9}=\frac{14!}{8!\;6!}\times 64\times 729\times a^{6}\;b^{8}\\\\\) \(\\\\\Rightarrow T_{9}=\frac{14\times 13\times 12\times 11\times 10\times 9\times 8! }{6\times 5\times 4\times 3\times 2\times 1\times 8!}\times 64\times 729\times a^{6}\;b^{8}\\\\\) \(\\\\\Rightarrow T_{9} = 3003\times 64\times 729\times a^{6}\;b^{8}=140107968\;\;a^{6}b^{8}\\\\\)

Therefore, the Coefficient of a⁶b⁸in the expansion of (2a – 3b)¹⁴= 140107968

Q.3: Write the general term in the expansion of (x³ – y²)⁵

Sol.

The general term in the expansion of (a + b)ⁿis given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

Now, on substituting n = 5, a = x³ and b = y² we will get:

T_r+1 = ⁵C_r× (x³)^{5 – r} × (y²)^r

Therefore, the general term in the expansion of (x³ – y²)⁵:

T_r+1 = ⁵C_r× (x)^{15 – 3 r} × (y)^{2 r}

Q.4: Write the general term in the expansion of (x⁴ – xy²)⁹

Sol.

The general term in the expansion of (a + b)ⁿis given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

Now, on substituting n = 9, a = x⁴ and b = xy² we will get:

T_r+1 = ⁹C_r× (x⁴)^{9 – r} × (xy²)^r

\(\Rightarrow\) T_r+1 = ⁹C_r× (x)^{36 – 4 r} × (x)^{2 r} × (y)^{2 r}

\(\Rightarrow\) T_r+1 = ⁹C_r× (x)^{36 – 4 r + 2 r} × (y)^{2 r}

\(\Rightarrow\) T_r+1 = ⁹C_r× (x)^{36 – 2 r}× (y)^{2 r}

Therefore, the general term in the expansion of (x⁴ – xy²)⁹:

T_r+1 = ⁹C_r× (x)^{36 – 2 r} × (y)^2r

Q.5: Find the 4^th term in the expansion of (2x + 3y)¹⁰

Sol.

Since, The general term in the expansion of (a + b)ⁿis given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

Therefore, from the above equation the value of r should be 3; for finding out the values of 4^th Term (T₄)

Therefore, T_{3 + 1} = ⁿC₃× (a)^{n – 3} × b³

Now, on substituting n =10, a = 2x and b = 3y we will get:

\(\\\Rightarrow\) T_{3 + 1} = ¹⁰C₃× (2x)^{10 – 3} × (3y)³

\(\\\Rightarrow\) T₄ = ¹⁰C₃× (2x)⁷ × (3y)³

\(\\\\\Rightarrow T_{4}=\frac{10!}{3!\; 7!}\times 128\;x^{7}\times 27\;y^{3}\\\\\) \(\\\\\Rightarrow T_{4}=\frac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}\times 3456\;x^{7}y^{3}\\\\\) \(\\\\\Rightarrow T_{4} = 120\times 3456\;x^{7}y^{3}=414720\;x^{7}y^{3}\\\\\)

Therefore, 4^th term in the expansion of (2x + 3y)¹⁰ = 414720 x⁷y³

Q.6 Find the 11^th term in the expansion of \(\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}\)

Sol.

Since, The general term in the expansion of (a + b)ⁿis given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

Therefore, from the above equation the value of r should be 10 for finding out the values of 11^th Term (T₁₁)

Therefore, T_{10 + 1} = ⁿC₁₀× (a)^{n – 10} × b¹⁰

Now, on substituting n =15, a = 8x and b = \(\left (-\frac{1}{2\sqrt{x}} \right )\) we will get:

\(\\\Rightarrow\) T_{10 + 1} = ¹⁵C₁₀× (8x)^{15 – 10} × \(\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\\)

\(\\\Rightarrow\) T₁₁ = ¹⁵C₁₀× (8x)⁵ × \(\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\\)

\(\\\\\Rightarrow T_{11}=\frac{15!}{10!\; 5!}\times 8\times 8\times 8\times 8\times 8\times x^{5}\times \frac{1}{2^{10}}\times \left ( \frac{1}{\sqrt{x}} \right )^{10}\\\\\) \(\\\\\\\Rightarrow T_{11}=\frac{15\times 14\times 13\times 12\times 11\times 10!}{5\times 4\times 3\times 2\times 1\times 10!}\times 32\times x^{5}\times \frac{1}{x^{5}}\\\\\) \(\\\\\Rightarrow T_{11} = 3003\times 32\boldsymbol{=96096}\\\\\)

Therefore, 11^th term in the expansion of \(\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}\)= 96096

Q.7: In the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\), Find the middle terms.

Sol.

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)ⁿ are given by:

middle terms = \(\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term\)

Therefore, the middle terms in the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\)are:

\(\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term\)

\(\Rightarrow\) 5^th term and 6^th term

Now, 5^th and 6^th terms in the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\) are:

T₅ = T_{4 + 1} = ⁹C₄× (5)^{9 – 4} × \(\left ( -\frac{x^{4}}{10} \right )^{4}\)

\(\\\\\Rightarrow\) T₅ = ⁹C₄× (5)⁵ × \(\left ( -\frac{x^{4}}{10} \right )^{4}\\\\\)

\(\\\\\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (5)^{5}\times \left ( -\frac{x^{4}}{10} \right )^{4}\\\\\) \(\\\\\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 5\times 5\times 5\times 5\times 5\times \frac{1}{10^{4}}\times x^{16}\\\\\) \(\\\\\Rightarrow T_{5} = 126\times \frac{5}{16}\;x^{16} \boldsymbol{=\frac{315}{8}\;x^{16}}\\\\\)

Therefore, 5^th term \(=\frac{315}{8}\;x^{16}\)

Now, T₆ = T_{5 + 1} = ⁹C₅× (5)^{9 – 5} × \(\left ( -\frac{x^{4}}{10} \right )^{5}\)

\(\\\\\Rightarrow\) T₆ = ⁹C₅× (5)⁴ × \(\left ( -\frac{x^{4}}{10} \right )^{4}\\\\\)

\(\\\\\Rightarrow T_{6}=\frac{9!}{5!\; 4!}\times (5)^{4}\times \left ( -\frac{x^{4}}{10} \right )^{5}\\\\\) \(\\\\\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 5\times 5\times 5\times 5\times \frac{-1}{10^{5}}\times x^{20}\\\\\) \(\\\\\Rightarrow T_{6} = -126\times \frac{1}{320}\;x^{20}\boldsymbol{=\frac{-63}{160}\;x^{20}}\\\\\)

Therefore, 6^th term \(=\frac{63}{160}\;x^{20}\)

Therefore, 5^th and 6^th terms are the middle terms in the expansion of \(\left ( 5-\frac{x^{4}}{10} \right )^{9}\)

And also, 5^th term \(=\frac{315}{8}\;x^{16}\) and 6^th term \(=\frac{-63}{160}\;x^{20}\)

Q.8: In the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\), Find the middle term.

Sol.

Here, n = 10

When n is even, the middle term in the expansion of (a + b)ⁿ is given by:

\(\left ( \frac{n}{2}+1 \right )^{th}term\)

Therefore, the middle term in the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\) is:

\(\Rightarrow \left ( \frac{10}{2}+1 \right )^{th}term\) = 6^th term

Now, 6^th term in the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\) is:

T₆ = T_{5 + 1} = ¹⁰C₅× \(\left ( \frac{x}{2} \right )^{10-5}\) × (8y)⁵

\(\\\Rightarrow\) T₆ = ¹⁰C₅× \(\left ( \frac{x}{2} \right )^{5}\) × (8y)⁵

\(\\\Rightarrow T_{6}=\frac{10!}{5!\; 5!}\times \left ( \frac{x}{2} \right )^{5}\times (8y)^{5}\\\) \(\\\Rightarrow T_{6}=\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}\times \frac{x^{5}}{32}\times 8\times 8\times 8\times 8\times 8\times y^{5}\\\) \(\\\Rightarrow T_{6} = 252\times 1024\;x^{5}y^{5}\boldsymbol{=258048\;x^{5}y^{5}}\\\)

Therefore, 6^th term = 258048 x⁵y⁵

Hence, the middle term in the expansion of \(\left ( \frac{x}{2}+8y \right )^{10}\) = 258048 x⁵y⁵

Q.9: Prove that the coefficients of p^aand p^b are equal in the expansion of (1 + p)^{a + b}

Sol.

Since, The general term in the expansion of (x + y)ⁿis given by: T_{r + 1} = ⁿC_r× (x)^{n – r} × (y)^r

Now, on substituting n = (a + b), x =1 and y = p, we will get:

T_{r + 1} = ^{(a + b)}C_r× (1)^{(a + b) – r} × (p)^r

\(\Rightarrow\) T_{r + 1}= ^{(a + b)}C_r × (p)^r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of p in equation (1) with p^a, we will get r = a

Therefore, from equation (1):

T_{a + 1} = ^{(a + b)}C_a × (p)^a

\(\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;(a+b-a)!}\times p^{a}\\\) \(\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;b!}\times p^{a}\\\)

Therefore, the coefficient of (p)^a = \(\frac{(a+b)!}{a!\;b!}\) . . . . . . . . . . (2)

Now, on comparing the coefficient of p in equation (1) with p^b, we will get r = b

Therefore, from equation (1):

T_{b + 1} = ^{(a + b)}C_b × (p)^b

\(\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;(a+b-b)!}\times p^{b}\\\\\) \(\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;a!}\times p^{b}\\\\\)

Therefore, the coefficient of (p)^b = \(\frac{(a+b)!}{b!\;a!}\) . . . . . . . . . . (3)

Now, on comparing equation (2) and equation (3) we conclude that the coefficients of p^aand p^b are equal in the expansion of (1 + p)^{a + b}.

Hence, Proved

Q.10: The coefficients of the (k + 1)^th, k^th and (k – 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 5 : 3 : 1. Find the values of n and k.

Sol.

Since, The general term in the expansion of (a + b)ⁿis given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

Now on substituting a = x and b = 1 in the above equation:

T_{r + 1} = ⁿC_r× (x)^{n – r} × (1)^r

Now, (k + 1)^th term in the expansion of (x + 1)ⁿ :

T_{(k) + 1} = ⁿC_k× (x)^{n – k} × 1^{(k )}

i.e. T_{(k + 1)} = ⁿC_k× (x)^{n – k}

\(\\\Rightarrow T_{k+1} = \frac{n!}{k!\;(n-k)!}\times (x)^{n-k}\\\)

Therefore the coefficient of T_{(k + 1)}^thterm = \(\frac{n!}{k!\;(n-k)!}\\\\\)

Now, (k)^th term in the expansion of (x + 1)ⁿ :

T_{(k – 1) + 1} = ⁿC_{k – 1}× (x)^{n – (k – 1)} × 1^{(k – 1)}

i.e. T_k = ⁿC_{k – 1} × (x)^{n – k + 1}

\(\\\\\Rightarrow T_{k} = \frac{n!}{(k-1)!\;(n-k+1)!}\times (x)^{n-k+1}\\\\\)

Therefore the coefficient of T_(k)^thterm = \(\frac{n!}{(k-1)!\;(n-k+1)!}\)

And, (k – 1)^th term in the expansion of (x + 1)ⁿ :

T_{(k – 2 )+ 1} = ⁿC_{k – 2}× (x)^{n – (k – 2)} × 1^{(k – 2)}

i.e. T_{k – 1} = ⁿC_{k – 2}× (x)^{n – k + 2}

\(\\\\\Rightarrow T_{k-1} = \frac{n!}{(k-2)!\;(n-k+2)!}\times (x)^{n-k+2}\\\\\)

Therefore the coefficient of T_{(k – 1)}^thterm = \(\frac{n!}{(k-2)!\;(n-k+2)!}\)

Now, according to the given conditions: Coefficients of T_{(k + 1)}, T_k and T_{k – 1} are in the ratio of 5 : 3 : 1

Therefore, \(\frac{T_{k-1}}{T_{k}}=\frac{1}{3}\\\)

\(\\\\\Rightarrow \left [ \frac{n!}{(k-2)!\;(n-k+2)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{1}{3}\\\\\) \(\\\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{(k-2)!\;(n-k+2)!}=\frac{1}{3}\\\\\) \(\\\\\Rightarrow \frac{(k-1)(k-2)!\;(n-k+1)!}{(k-2)!\;(n-k+2)(n-k+1)!}=\frac{1}{3}\\\\\) \(\\\Rightarrow \frac{k-1}{n-k+2}=\frac{1}{3}\\\) \(\\\Rightarrow 3k-3=n-k+2\\\)

(or) n – 4k + 5 = 0 . . . . . . . . . . . . . . . . (1)

And, \(\frac{T_{k+1}}{T_{k}}=\frac{5}{3}\\\)

\(\\\Rightarrow \left [ \frac{n!}{k!\;(n-k)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{5}{3}\\\) \(\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{k!\;(n-k)!}=\frac{5}{3}\\\) \(\\\Rightarrow \frac{(k-1)!\;[(n-k+1)\;(n-k)!]}{[k\;(k-1)!]\;(n-k)!}=\frac{5}{3}\\\) \(\\\Rightarrow \frac{n-k+1}{k}=\frac{5}{3}\\\) \(\\\Rightarrow 3n-3k+3=5k\\\)

(or) 3n – 8k + 3 = 0 . . . . . . . . . . . . . . . . . . (2)

On multiplying equation (1) by 3 and subtracting it to equation (2) we will get:

(3n – 8k + 3) – 3(n – 4k + 5) = 0

\(\\\Rightarrow\) 3n – 8k + 3 – 3n + 12k – 15 = 0

\(\\\Rightarrow\) 4k = 12

Therefore, k = 3

Now on substituting the value of k in equation (2) we will get:

3n – 8(3) + 3 = 0

Therefore, n = 7

Q.11: Prove that the coefficient of a^p in the expansion of (1 + 3a)^{2p – 1}is half the coefficient of a^p in the expansion of (1 + 3a)^2p

Sol.

Since, The general term in the expansion of (x + y)ⁿis given by: T_{r + 1} = ⁿC_r× (x)^{n – r} × (y)^r

For (1 + 3a)^2p:

On putting n = 2p, x =1 and y = 3a, we will get:

T_{r + 1} = ^(2p)C_r× (1)^{(3p) – r} × (3a)^r

\(\\\\\Rightarrow\) T_{r + 1}= ^(2p)C_r × (3a)^r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of a in equation (1) with a^p, we will get r = p

Therefore, from equation (1):

T_{p + 1} = ^(2p)C_p × (3a)^p

\(\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;(2p-p)!}\times (3a)^{p}\\\\\)

\(\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;p!}\times 3^{p}\times (a)^{p}\\\\\)

Therefore, the coefficient of a^p \(=\frac{(2p)!}{p!\;p!}\times 3^{p}\) . . . . . . . . . . . . . . . . (2)

Now, for (1 + 3a)^{2p – 1}:

On substituting n = (2p – 1), x =1 and y = 3a, we will get:

T_{r + 1} = ^{(2p – 1)}C_r× (1)^{(2p – 1) – r} × (3a)^r

\(\Rightarrow\) T_{r + 1}= ^{(2p – 1)}C_r × (3a)^r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of ‘a’ in equation (1) with ‘a^p’, we will get r = p

Therefore, from equation (1):

T_{p + 1} = ^{(2p – 1)}C_p × (3a)^p

\(\\\\\Rightarrow T_{p+1}=\frac{(2p-1)!}{p!\;(2p-1-p)!}\times (3a)^{p}\\\\\) \(\\\\\Rightarrow T_{p+1}= \frac{(2p-1)!}{p!\;(p-1)!}\times (3)^{p}\times (a)^{p}\\\\\) \(\\\\\Rightarrow T_{p+1}=\frac{\frac{(2p)!}{2p}}{p!\times \frac{p!}{p}}\times (3)^{p}.(a)^{p}=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}.a^{p}\\\)

Therefore, the coefficient of a^p = \(=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}\) . . . . . . . . . . (3)

Now, on comparing equation (2) and equation (3):

\(\\\\\Rightarrow \frac{(2p)!}{p!\;p!}\times 3^{p}=\frac{(2p)!}{p!\;p!}\times 3^{p}\\\\\)

\(\\\Rightarrow\) ^(2p)C_p × 3^p= \(\frac{1}{2}\) ^{(2p – 1)}C_p × 3^p

Therefore, the coefficient of a^p in the expansion of (1 + 3a)^{2p – 1} is half the coefficient of a^p in the expansion of (1 + 3a)^2p

Q.12: For what values of ‘m’ the coefficient of x² in the expansion of (1+ 2x)^k is 140.

Sol.

Since, The general term in the expansion of (a + b)ⁿis given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × (b)^r

Now, on substituting n = m, a =1 and b = 2x we will get:

T_{r + 1} = ^mC_r× (1)^{m – r} × (2x)^r

\(\Rightarrow\) T_{r + 1}= ^mC_r × (2)^r × (x)^r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x², we will get r = 2

Therefore, from equation (1):

T_{2 + 1} = ^mC₂ × (2)² × (x)²

Since, the coefficient of x² = 140 [GIVEN]

Therefore, 140 = ^mC₂ × (2)²

\(\Rightarrow 84=\left [ \frac{m!}{2!\;\;(m-2)!} \right ]\times 4\\\\\) \(\Rightarrow 21=\frac{m\times (m-1)\;\;(m-2)!}{[2\times 1]\times (m-2)!}\\\\\) \(\Rightarrow 42 = m^{2}-m\)

Therefore, m² – m – 42 = 0

Now, by splitting of middle term method, the roots of this quadratic equation are:

\(\Rightarrow\)m² – (7 – 6)m – 42 = 0

\(\Rightarrow\)m² – 7m + 6m – 42 = 0

\(\Rightarrow\) m(m – 7) +6(m – 7) = 0

i.e. (m + 6) (m – 7) = 0

Therefore, m = 7 or m = -6

Hence, for the coefficient of x² in the expansion of (1+ 2x)^k to be 140, the value of ‘m’ should be 7.

Q.13: In the expansion of (2x+3y)⁹, Find the middle terms.

Sol.

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)ⁿ are given by:

middle terms = \(\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term\)

Therefore, the middle terms in the expansion of (2x+3y)⁹ are:

\(\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term\)

\(\Rightarrow\) 5^th term and 6^th term

Now, 5^th and 6^th terms in the expansion of (2x+3y)⁹ are:

T₅ = T_{4 + 1} = ⁹C₄× (2x)^{9 – 4} × (3y)⁴

\(\Rightarrow\) T₅ = ⁹C₄× (2x)⁵ × (3y)⁴

\(\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (2x)^{5}\times (3y)^{4}\) \(\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 32x^{5}\times 81y^{4}\\\\\)

\(\Rightarrow T_{5} = 126\times 2592\;x^{5}y^{4}\boldsymbol{=414720\;x^{5}y^{4}}\\\)

Therefore, 5^th term = 414720 x⁵ y⁴

Now, T₆ = T_{5 + 1} = ⁹C₅× (2x)^{9 – 5} × (3y)⁵

\(\Rightarrow\) T₆ = ⁹C₅× (2x)⁴ × (3y)⁵

\(\Rightarrow T_{6}=\frac{9!}{5!\times 4!}\times (2x)^{4}\times (3y)^{5}\) \(\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 16x^{4}\times 243y^{5}\\\\\) \(\Rightarrow T_{6} = 126\times 3888\;x^{4}y^{5}\boldsymbol{=489888\;x^{4}y^{5}}\\\)

Therefore, 6^th term = 489888 x⁴ y⁵

Hence, 5^th and 6^th terms are the middle terms in the expansion of (2x+3y)⁹

And also, 5^th term = 414720 x⁵ y⁴ and 6^th term = 489888 x⁴ y⁵

Q.14: Find the Coefficient of x⁶in expansion of (x + 3)¹¹

Sol.

Since, The general term in the expansion of (a + b)ⁿis given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

Therefore, on substituting n = 11, a = x and b = 3 in the above expression we will get:

T_{r + 1} = ¹¹C_r× (x)^{11 – r} × 3^r . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x⁷, we will get:

i.e. (x)^{11 – r} = x⁶

(or) 11 – r = 6

Therefore, r = 5

Now, on substituting r in equation (1) we will get:

T_{5 + 1} = ¹¹C₅× (x)^{11 – 5} × 3⁵

\(\Rightarrow T_{6}=\frac{11!}{5!\times 6!}\times x^{6}\times 3^{5}\) \(\Rightarrow T_{6}=\frac{11\times 10\times 9\times 8\times 7\times 6!}{5\times 4\times 3\times 2\times 1\times 6!}\times 243\;x^{6}\\\\\)

\(\Rightarrow T_{6} = 462\times 243\;x^{6}\boldsymbol{=112266\;x^{6}}\)

Therefore, the Coefficient of x⁶in the expansion of (x + 3)¹¹ = 112266

Q.15: Write the general term in the expansion of (x²y³ – x³ y²)⁷

Sol.

The general term in the expansion of (a + b)ⁿis given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

Now, on substituting n = 7, a = x²y³ and b = x³ y² we will get:

T_r+1 = ⁷C_r× (x²y³)^{7 – r} × (x³ y²)^r

\( \Rightarrow\) T_{r + 1} = ⁷C_r× (x²)^{7 – r} × (y³)^{7 – r} × (x)^{3 r}× (y)^{2 r}

\( \Rightarrow\) T_{r + 1} = ⁷C_r× (x)^{14 – 2 r} × (x)^{3 r}× (y)^{2 r}× (y)^{21 – 3r}

\( \Rightarrow\) T_{r + 1} = ⁷C_r× (x)^{14 – 2 r + 3r}× (y)^{21 – 3r + 2 r}

i.e. T_{r + 1} = ⁷C_r× (x)^{14 + r}× (y)^{21 – r}

Therefore, the general term in the expansion of (x²y³ – x³ y²)⁷:

T_{r + 1} = ⁷C_r× (x)^{14 + r} × (y)^{21 – r}

Miscellaneous Exercise:

Q.1: If 1^st term, 2^nd term and 3^rd term in the expansion of (a + b)ⁿ are 729, 7290 and 30375 respectively then find the value of ‘a’, ‘b’ and ‘c’.

Sol.

The general term in the expansion of (a + b)ⁿ is given by: T_{r + 1} = ⁿC_r× (a)^{n – r} × b^r

Now, according to the given conditions:

T_{(0 + 1)} = ⁿC₀× (a)^{n – 0} × b⁰

T₁ = aⁿ

i.e. aⁿ = 729 . . . . . . . . . . . . . . . . . . (1)

Also, T_{(1 + 1)} = ⁿC₁× (a)^{n – 1} × b¹

T₂ = ⁿC₁× (a)^{n – 1} × b

i.e. 7290 = ⁿC₁× (a)^{n – 1} × b

\(\\\Rightarrow 7290=\frac{n!}{1!\;\;(n-1)!}\times a^{n}\times a^{-1}\times b\\\\\) \(\\\\\Rightarrow 7290=\frac{n(n-1)!}{(n-1)!}\times a^{n}\times \frac{b}{a}\;\;\;\)

Therefore, 7290 = n × aⁿ × \(\frac{b}{a}\) . . . . . . . . . . . . . . . . (2)

And, T_{(2 + 1)} = ⁿC₂× (a)^{n – 2} × b²

T₃ = ⁿC₂× (a)^{n – 2} × b²

i.e. 30375 = ⁿC₂× (a)^{n – 2} × b²

\(\\\Rightarrow 30375=\frac{n!}{2!\;\;(n-2)!}\times a^{n}\times a^{-2}\times b^{2}\\\) \(\\\Rightarrow 30375=\frac{n\;(n-1)\;(n-2)!}{(2\times 1)\;\;(n-2)!}\times a^{n}\times \left ( \frac{b}{a} \right )^{2}\\\)

Therefore, 60750 = n (n – 1) × aⁿ× \(\left ( \frac{b}{a} \right )^{2}\) . . . . . . . . . . . . . . . . (3)

Now, on substituting equation (1) in equation (2) we will get:

\(\\\Rightarrow 7290 = n\times (729)\times \frac{b}{a}\\\) \(\\\Rightarrow \frac{10a}{b}=n\\\)

Therefore, n = \(\frac{10a}{b}\) . . . . . . . . . . . . . . . . . (4)

Now, on dividing equation (2) and equation (3) we will get:

\(\\\Rightarrow \frac{n \times a^{n} \times \frac{b}{a}}{n\times (n-1)\times a^{n}\times \left ( \frac{b}{a} \right )^{2} }=\frac{7290}{60750}\\\) \(\\\Rightarrow \frac{3}{25}=\frac{1}{(n-1)\times \frac{b}{a}}\\\) \(\\\Rightarrow 3n-3=\frac{25a}{b}\\\)

Now, from equation (4):

\(\frac{a}{b}\) = \(\frac{n}{10}\)

\(\Rightarrow 3n-3=25\times \left ( \frac{n}{10} \right )\) \(\Rightarrow 30n-30=25n\)

i.e. 30n – 30 = 25n

Therefore, n = 6

Now, on substituting the value of ‘n’ in equation (1) we will get:

\(\Rightarrow\) a⁶ = 729

i.e. a⁶ = 3⁶

Therefore, a = 3

Now, from equation (4):

\(\\\Rightarrow\) 6 = \(\frac{30}{b}\\\)

Therefore, b = 5

Q.2: If the coefficients of z² and z³ are equal in the expansion of (3 + bz)⁹. Find the value of ‘b’

Sol.

Since, The general term in the expansion of (x + y)ⁿ is given by: T_{r + 1} = ⁿC_r× (x)^{n –}^r × y^r

Therefore, on substituting n = 9, x = 3 and y = bz in the above expression we will get:

T_{r + 1} = ⁹C_r× (3)^{9 – r} × (bz)^r

T_{r + 1} = ⁹C_r× (3)^{9 – r} × b^r × z^r . . . . . (1)

Now, on comparing the coefficient of z in equation (1) with z², we will get: r = 2

On substituting the value of r in equation (1) we will get:

T_{(2 + 1)} = ⁹C₂× (3)^{9 – 2} × b² × z²

\(\\\Rightarrow T_{3}=\frac{9!}{2!\;7!}\times 3^{7}\times b^{2}\times z^{2}\\\) \(\\\Rightarrow T_{3}=\frac{9\times 8\times 7!}{2\times 1\times 7!}\times 3^{7}\times b^{2}\times z^{2}\\\)

\(\\ \Rightarrow T_{3} = 36\times 3^{7}\times b^{2}\times z^{2}\) . . . . . . . . . . . . (2)

Now, on comparing the coefficient of z in equation (1) with z³, we will get: r = 3

On substituting the value of ‘r’ in equation (1) we will get:

T_{(3 + 1)} = ⁹C₃× (3)^{9 – 3} × b³ × z³

\(\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times 3^{6}\times b^{3}\times z^{3}\\\) \(\\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 3^{6}\times b^{3}\times z^{3}\\\)

\(\\\Rightarrow T_{4} = 84\times 3^{6}\times b^{3}\times z^{3} . . . . . . . . . (3)\)

Now, according to the given conditions:

Coefficient of z² = Coefficient of z³

Therefore from equation (2) and equation (3):

\(\\\Rightarrow 36\times 3^{7}\times b^{2}=84\times 3^{6}\times b^{3}\\\) \(\\\Rightarrow b = \frac{36\times 3^{7}}{84\times 3^{6}}\\\)

Therefore, the value of b = \(\frac{9}{7}\)

Q.3: Find the coefficient of x⁶ in the product of (1 + 3x)⁷ (1 – 2x)⁶ by using binomial theorem.

Sol.

By using Binomial Theorem:

Since, (1 + x)ⁿ = [ ⁿC₀ ] + [ ⁿC₁× (x) ] + [ ⁿC₂× (x)²] + [ ⁿC₃× (x³) ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n× (x)ⁿ]

Therefore, (1 + 3x)⁷ = [ ⁷C₀] + [ ⁷C₁× (3x) ] + [ ⁷C₂× (3x)²] + [ ⁷C₃× (3x)³] + [ ⁷C₄× (3x)⁴ ] + [ ⁷C₅× (3x⁵) ] + [ ⁷C₆× (3x)⁶ ] + [ ⁷C₇× (3x)⁷ ]

\(\\\Rightarrow\) 1 + [ 7 × (3x) ] + [ 21 × (9x²) ] + [ 35 × (27x³) ] + [ 35 × (81x⁴) ] + [ 21 × (243x⁵) ] + [ 7 × 729x⁶ ] + [ 1 × 2187x⁷ ]

\(\\\Rightarrow\) 1 + 21x + 189x² + 945x³ + 2835x⁴ + 5103x⁵+ 5103x⁶+ 2187x⁷

Therefore, (1 + 3x)⁷ = 1 + 21x + 189x² + 945x³ + 2835x⁴ + 5103x⁵+ 5103x⁶+ 2187x⁷

Now, (1 – 2x)⁶:

By using Binomial Theorem:

(1 – 2x)⁶ = [ ⁶C₀] – [ ⁶C₁× (2x) ] + [ ⁶C₂× (2x)²] – [ ⁶C₃× (2x)³] + [ ⁶C₄× (2x)⁴ ] – [ ⁶C₅× (2x⁵) ] + [ ⁶C₆× (2x)⁶ ]

\(\\\Rightarrow\) 1 – [ 6 × (2x) ] + [ 15 × (4x²) ] – [ 20 × (8x³) ] + [ 15 × (16x⁴) ] – [ 6 × (32x⁵) ] + [ 1 × 64x⁶ ]

\(\\\Rightarrow\) 1 – 12x + 60x² – 160x³ + 240x⁴ – 192x⁵+ 64x⁶

Therefore, (1 – 2x)⁶= 1 – 12x + 60x² – 160x³ + 240x⁴ – 192x⁵+ 64x⁶

Now, (1 + 3x)⁷ (1 – 2x)⁶:

=(1 + 21x + 189x² + 945x³ + 2835x⁴ + 5103x⁵+ 5103x⁶+ 2187x⁷) × (1 – 12x + 60x² – 160x³ + 240x⁴ – 192x⁵+ 64x⁶)

Now, it not required to multiply the above equation, only terms with x⁵ are required and that can be identified by analyzing the above equation.

\(\\\Rightarrow\) [{1 × (- 192x⁵)} + {(21x) × (240x⁴)} + {(189x²) × (- 160x³)} + {(945x³) × (60x²)} + {(2835x⁴) × (- 12x)} + {(5103x⁵) × (1)}]

\(\\\Rightarrow\) [ -192x⁵ + 5040x⁵ – 30240x⁵+ 56700x⁵– 34020x⁵ + 5103x⁵]

\(\\\Rightarrow\) 2391 x⁵

Therefore, the coefficient of x⁵ = 2391

Q.4: Show that (a – b) is a factor of (aⁿ – bⁿ), where ‘n’ is a positive integer and ‘a’ and ‘b’ are distinct integers.

Sol.

Now, for (a – b) to be factor of (aⁿ – bⁿ), (aⁿ – bⁿ) = p(a – b) [where p is any natural number]

We can write aⁿ = (a – b + b)ⁿ = [(a – b) + b]ⁿ

Now, by Binomial Theorem:

Therefore, for x = (a – b) and y = b the equation becomes:

[(a – b) + b]ⁿ = [ ⁿC₀× (a – b)ⁿ ] + [ ⁿC₁× (a – b)^{n – 1} × b ] + [ ⁿC₂× (a – b)^{n – 2} × b²] + [ ⁿC₃× (a – b)^{n – 3}× b³] + . . . . . . . . . . . . . + [ ⁿC_{n – 1}× (a – b) × b^{n – 1}] + [ ⁿC_n× bⁿ]

\(\\\Rightarrow\) [a]ⁿ = [(a – b)ⁿ] + [ ⁿC₁× (a – b)^{n – 1} × b ] + [ ⁿC₂× (a – b)^{n – 2} × b²] + [ ⁿC₃× (a – b)^{n – 3}× b³] + . . . . . . . . . . . . . + [ ⁿC_{n – 1}× (a – b) × b^{n – 1}] + [ bⁿ]

Now, aⁿ – bⁿ = [(a – b)ⁿ] + [ ⁿC₁× (a – b)^{n – 1} × b ] + [ ⁿC₂× (a – b)^{n – 2} × b²] + [ ⁿC₃× (a – b)^{n – 3}× b³] + . . . . . . . . . . . . . + [ ⁿC_{n – 1}× (a – b) × b^{n – 1}] + [ bⁿ] – [ bⁿ]

\(\\\Rightarrow\) aⁿ – bⁿ = (a – b) × [ (a – b)^{n – 1} + ⁿC₁(a – b)^{n – 2} b + ⁿC₂(a – b)^{n – 3} b²+ ⁿC₃(a – b)^{n – 4}b³+ . . . . . . . . . . . . . + ⁿC_{n – 1} b^{n – 1}]

\(\\\Rightarrow\) (aⁿ – bⁿ) = (a – b) × p

Where, p = [ (a – b)^{n – 1} + ⁿC₁(a – b)^{n – 2} b + ⁿC₂(a – b)^{n – 3} b²+ ⁿC₃(a – b)^{n – 4}b³+ . . . . . . . . . . . . . + ⁿC_{n – 1} b^{n – 1}] is any natural number.

Therefore, (a – b) is a factor of (aⁿ – bⁿ), where ‘n’ is a positive integer.

Q.5: Evaluate \(\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}\)

Sol.

Find (a + b)⁶ – (a – b)⁶

Now, by using Binomial Theorem:

Therefore, [a + b]⁶= [ ⁶C₀× (a⁶) ] + [ ⁶C₁× (a⁵) × b ] + [ ⁶C₂× (a⁴) × b²] + [ ⁶C₃× (a³) × b³] + [ ⁶C₄× (a²) × b⁴] + [ ⁶C₅× a × b⁵] + [⁶C₆ × b⁶]

And, [a – b]⁶ = [ ⁶C₀× (a⁶) ] – [ ⁶C₁× (a⁵) × b ] + [ ⁶C₂× (a⁴) × b²] – [ ⁶C₃× (a³) × b³] + [ ⁶C₄× (a²) × b⁴] – [ ⁶C₅× a × b⁵] + [⁶C₆ × b⁶]

\(\\\Rightarrow\) 2[ ⁶C₁a⁵b + ⁶C₃a³b³+⁶C₅ab⁵] = 2[ 6a⁵b + 20a³b³+ 6ab⁵ ]

Therefore, (a + b)⁶– (a – b)⁶ = 4ab[ 3a⁴ + 10a²b² + 3b⁴] . . . . . . (1)

Now, on substituting a = \(\sqrt{7}\) and b = \(\sqrt{5}\) in equation (1) we will get:

\(\\\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}:\\\)

\(\\=4( \sqrt{7}\times \sqrt{5} )\left [ (3 \sqrt{7} )^{4} +10\times (\sqrt{7} )^{2}\times ( \sqrt{5})^{2}+3(\sqrt{5})^{4} \right ]\\\)

\(\\\Rightarrow 4\sqrt{35}\;[\;147+350+75\;]\\\)

Therefore, \(\left(\sqrt{7}+\sqrt{5} \right )^{6}-\left ( \sqrt{7}-\sqrt{5}\right)^{6}\)= \(2288\sqrt{35}\)

Q.6: Find the expansion of \(\left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}\)

Sol.

The above given expression can be assumed as: (x + y)⁴ + (x – y)⁴

Now, by using Binomial Theorem:

Therefore, [x + y]⁴= [⁴C₀× (x⁴)] + [⁴C₁× (x³) × (y)] + [⁴C₂× (x²) × (y)²] + [⁴C₃× (x¹) × (y)³] + [⁴C₄× (x⁰) × (y)⁴]

And, [x – y]⁴ = [⁴C₀× (x⁴) ] – [⁴C₁× (x³) × (y) ] + [⁴C₂× (x²) × (y)²] – [⁴C₃× (x¹) × (y)³] + [⁴C₄× (x⁰) × (y)⁴]

Therefore, (x + y)⁴+ (x – y)⁴ = [ ⁴C₀x⁴ + ⁴C₁x³y + ⁴C₂x²y²+ ⁴C₃x y³+ ⁴C₄y⁴] + [ ⁴C₀x⁴ – ⁴C₁x³y + ⁴C₂x²y² – ⁴C₃x y³+ ⁴C₄y⁴]

\(\\\Rightarrow\) [ ⁴C₀x⁴ + ⁴C₁x³y + ⁴C₂x²y²+ ⁴C₃xy³+ ⁴C₄y⁴ + ⁴C₀x⁴ – ⁴C₁x³y + ⁴C₂x²y² – ⁴C₃xy³+ ⁴C₄y⁴]

\(\\\Rightarrow\) 2[ ⁴C₀x⁴ + ⁴C₂x²y² + ⁴C₄y⁴] = 2[ x⁴ + 6x²y² + y⁴]

Therefore, (x + y)⁴+ (x – y)⁴ = 2[ x⁴ + y⁴+ 6 x²y²] . . . . . . . . . . . . . . . . . . . (1)

Now, on substituting x = a³ and y = \(\sqrt{a^{3}-2}\) in equation (1) we will get:

\(\\\Rightarrow \left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}:\\\)

\(\\=2\times \left [ (a^{3})^{4}+ (\sqrt{a^{3}-2})^{4}+[6\times (a^{3})^{2}\times (\sqrt{a^{3}-2})^{2}] \right ]\\\)

\(\\=2\left [ a^{12}+ (a^{3}-2)^{2}+[6\times a^{6}\times (a^{3}-2)] \right ]\\\)

\(\\=2\left [\; a^{12}+ a^{6}+4-4a^{3}+6a^{9}-12a^{6} \;\right ]\\\)

\(\\\Rightarrow 2a^{12}+12a^{9}-22a^{6}-8a^{3}+8\)

Therefore, the expansion of \(\left ( a^{3} +\sqrt{a^{3}-2}\right )^{4}+ \left ( a^{3} -\sqrt{a^{3}-2}\right )^{4}\):

= \(2a^{12}+12a^{9}-22a^{6}-8a^{3}+8\)

Q.7: Find the approximate value of (0.98)⁵ using the first four terms of its expansion.

Sol.

(0.98)⁵ = (1 – 0.02)⁵

Now, by using Binomial Theorem:

Since, [ x – y ]ⁿ = [ ⁿC₀× (x)ⁿ ] – [ ⁿC₁× (x)^{n – 1} × y ] + [ ⁿC₂× (x)^{n – 2} × y²] – [ ⁿC₃× (x)^{n – 3}× y³] + . . . . . . . . . . . . . . . . . . (-1)ⁿ[ ⁿC_n× yⁿ]

Therefore, (1 – 0.02)⁵ = [ ⁵C₀× (1)⁵ ] – [ ⁵C₁× (1)⁴ × (0.02) ] + [ ⁵C₂× (1)³ × (0.02)²] – [ ⁵C₃× (1)²× (0.02)³]

= 1 – [5 × 1 × 0.02] + [5 × 1 × 0.0004] – [5 × 1 × 0.000008]

= 1 – 0.10 + 0.0020 + 0.000040

= 0.90204

Therefore, (0.98)⁵ = 0.90204

Q.8: The coefficient of the 5^th term from end and the 5^th term from the beginning in the expansion of \(\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}\) are in the ratio \(1:\sqrt{6}\).Find the value of ‘n’.

Sol.

Since, The general term in the expansion of (a + b)ⁿis given by: T_{k + 1} = ⁿC_k× (a)^{n – k} × b^k

Now on substituting a = \(\left ( 2\right )^{{\frac{1}{4}}}\) and b = \(\frac{1}{(3)^{\frac{1}{4}}}\) in the above equation:

T_{k + 1} = ⁿC_k×\(\left ( \sqrt[4]{2} \right )^{n-k} \times \left ( \frac{1}{\sqrt[4]{3}}\right )^{k}\\\)

Now, 5^th term from beginning in the expansion of \(\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}\):

T_{(4 + 1)} = ⁿC₄× \(\left ( \sqrt[4]{2} \right )^{n-4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}\\\)

i.e. T₅ = ⁿC₄× \(\left ( 2 \right )^{\frac{n}{4}-1} \times \left ( \frac{1}{3} \right )\\\)

\(\\\Rightarrow T_{5} = \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times (2)^{-1}\times \frac{1}{3}\\\)

Therefore the coefficient of 5^thterm from beginning \(\boldsymbol{= \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6}}\\\)

Now, 5^th term from end in the expansion of \(\left ( 2^{\frac{1}{4}}+\frac{1}{3^{\frac{1}{4}}} \right )^{n}\):

1^st term, 2^nd term, 3^rd term . . . . . . . . . . . (n – 5)^th term, (n – 4)^th term, (n – 3)^th term, (n – 2)^th term, (n – 1)^th term, n^th term.

Therefore, 5^th term from end will be (n – 4)^th term from beginning.

i.e. T_{(n – 4) + 1} = ⁿC_{n – 4}× \(\left ( \sqrt[4]{2} \right )^{n-(n-4)} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}\\\)

i.e. T_{(n – 3)} = ⁿC_{n – 4} × \(\left ( \sqrt[4]{2} \right )^{4} \times \left ( \frac{1}{\sqrt[4]{3}} \right )^{n}\times \left ( \frac{1}{\sqrt[4]{3}} \right )^{-4}\\\)

\(\\\Rightarrow T_{(n-3)} = \frac{n!}{(n-4)!\;[(n-(n-4)]!}\times 2 \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 3\\\)

Therefore the coefficient of 5^th term from end [ T_{(n – 4)}^thterm ] \(= \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6\\\)

Now, according to the given conditions:

The ratio of coefficient of 5^th term from end and the coefficient of 5^thterm from beginning is \(1:\sqrt{6}\\\)

Therefore, \(\\ \frac{T_{n-3}}{T_{5}}=\frac{1}{\sqrt{6}}\\\)

\(\\\Rightarrow \left ( \frac{n!}{(n-4)!\;[(4)]!} \times \left ( \frac{1}{3} \right )^{\frac{n}{4}}\times 6 \right )\div\left ( \frac{n!}{4!\;(n-4)!}\times (2)^{\frac{n}{4}}\times \frac{1}{6} \right )=\frac{1}{\sqrt{6}}\\\) \(\\\Rightarrow \frac{4!\;(n-4)!}{(n-4)!\;4!}\times \frac{1}{(2)^{\frac{n}{4}}}\times \frac{1}{(3)^{\frac{n}{4}}}\times 6\times 6=\frac{1}{\sqrt{6}}\\\) \(\\\Rightarrow \frac{1}{(6)^{\frac{n}{4}}}\times 36=\frac{1}{\sqrt{6}}\\\) \(\\\Rightarrow 36\sqrt{6}=(6)^{\frac{n}{4}}\\\) \(\\\Rightarrow (6)^{2}\times (6)^{\frac{1}{2}}=6^{\frac{n}{4}}\\\) \(\\\Rightarrow (6)^{\frac{5}{2}}=(6)^{\frac{n}{4}\\}\)

On comparing RHS and LHS:

\(\\\Rightarrow \frac{5}{2}=\frac{n}{4}\\\)

\(\\\Rightarrow n = 10\)

Therefore, the value of n = 10.

Q.9: By using the Binomial Theorem, find the expansion of \(\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}\); where ≠ 0

Sol.

The given expression \(\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}\) can be expanded as \(\left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4}\)

Now, by using Binomial Theorem:

Therefore, \(\left [\left (1-\frac{3}{x} \right )+\frac{x}{3} \right ]^{4}\) = [⁴C₀× \(\left (1-\frac{3}{x} \right )^{4}\)] + [⁴C₁× \(\left (1-\frac{3}{x} \right )^{3}\) × \(\left ( \frac{x}{3} \right )\)] + [⁴C₂× \(\left (1-\frac{3}{x} \right )^{2}\) × \(\left ( \frac{x}{3} \right )^{2}\)] + [⁴C₃× \(\left (1-\frac{3}{x} \right )^{1}\) × \(\left ( \frac{x}{3} \right )^{3}\) + [⁴C₄× \(\left ( \frac{x}{3} \right )^{4}\)]

\(\\=\left [ 1\times \left (1-\frac{3}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{3}\times \left ( \frac{x}{3} \right ) \right ]+\left [ 6\times \left (1-\frac{3}{x} \right )^{2} \times \left ( \frac{x}{3} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{3}{x} \right )^{1} \times \left ( \frac{x}{3} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{3} \right )^{4} \right ]\) . . . . . . . . . . . . . . . . . . . (1)

Now, \(\left (1-\frac{3}{x} \right )^{4}\):

By using Binomial Theorem:

Since, [x – y]ⁿ = [ ⁿC₀× (xⁿ) ] – [ ⁿC₁× (x^{n – 1}) × y ] + [ ⁿC₂× (x^{n – 2}) × y²] – [ ⁿC₃× (x^{n – 3}) × y³] + . . . . . . . . . . . . . . . . . . (-1)ⁿ[ ⁿC_n× yⁿ]

Therefore, \(\left (1-\frac{3}{x} \right )^{4}\) = [⁴C₀] – [⁴C₁ × \(\left ( \frac{3}{x} \right )^{1}\)] + [⁴C₂ × \(\left ( \frac{3}{x} \right )^{2}\)] – [⁴C₃ × \(\left ( \frac{3}{x} \right )^{3}\)] + [⁴C₄ × \(\left ( \frac{3}{x} \right )^{4}\)]

\(\\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{3}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{27}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{81}{x^{4}} \right ) \right ]\\\)

\(\\=\left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}} \right ]\) . . . . . . . . (2)

And, \(\left (1-\frac{3}{x} \right )^{3}\) = [ ³C₀] – [ ³C₁ × \(\left ( \frac{3}{x} \right )^{1}\)] + [³C₂ × \(\left ( \frac{3}{x} \right )^{2}\)] – [³C₃ × \(\left ( \frac{3}{x} \right )^{3}\)]

\(\\= [1]-\left [ 3\times \left ( \frac{3}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{9}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{27}{x^{3}} \right ) \right ]\\\)

\(\\=\left [1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right ]\) . . . . . . . . . . . . (3)

Now, from equation (1), equation (2) and equation (3):

\(\\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ 4 \times \left ( 1-\frac{9}{x}+\frac{27}{x^{2}}-\frac{27}{x^{3}} \right )\times \frac{x}{3} \right ]+\left [ 6\times \left (1+\frac{9}{x^{2}}-\frac{6}{x} \right )\times \frac{x^{2}}{9}\right ] +\left [ \left ( 4-\frac{12}{x} \right )\times \frac{x^{3}}{27} \right ]+\left [\frac{x^{4}}{81}\right ]\\\) \(\\\Rightarrow \left [ 1-\frac{12}{x}+\frac{54}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]+\left [ \frac{4x}{3}-12+\frac{36}{x}-\frac{36}{x^{2}}\right ]+\\\\\\\\+\left [ \frac{2x^{2}}{3}+6-4x\right ] +\left [ \frac{4x^{3}}{27}-\frac{4x^{2}}{9} \right ]+\left [\frac{x^{4}}{81}\right ]\\\) \(\\\boldsymbol{\Rightarrow \left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}\\\)

Therefore, the expansion of \(\left ( 1+\frac{x}{3}-\frac{3}{x} \right )^{4}\):

= \(\boldsymbol{\left [ \frac{x^{4}}{81}+\frac{4x^{3}}{27}+\frac{2x^{2}}{9}-\frac{8x}{3}-5+\frac{24}{x}+\frac{18}{x^{2}}-\frac{108}{x^{3}}+\frac{81}{x^{4}}\right ]}\)

Q.10: By using the Binomial Theorem, find the expansion of (6x² – 4ax + 6a²)³.

Sol.

The given expression (6x² – 4ax + 6a²)³ can be expanded as [(6x² – 4ax) + 6a²)³]

By using Binomial Theorem:

Therefore, [(6x² – 4ax) + 6a²)³] = [³C₀× (6x² – 4ax)³] + [³C₁× (6x² – 4ax)² × 6a²] + [³C₂× (6x² – 4ax)¹ × (6a²)²] + [³C₃× (6x² – 4ax)⁰× (6a²)³]

= [1× (6x² – 4ax)³] + [3 × (6x² – 4ax)² × 6a²] + [3 × (6x² – 4ax) × (6a²)²] + [1 × (6a²)³]

= [(6x² – 4ax)³] + [(6x² – 4ax)² × 18a²] + [(6x² – 4ax) × 108a⁴] + [216a⁶] . . . . . . . . . . . . (1)

Now, [(6x² – 4ax)³]:

By using Binomial Theorem:

Therefore, [(6x² – 4ax)³] = [³C₀× (6x²)³] – [³C₁× (6x²)²) × 4ax] + [³C₂ × 6x² × (4ax)²] – [³C₃ × (4ax)³]

= [1 × (216x⁶)] – [3 × (36x⁴) × 4ax] + [3 × (6x²)× 16a²x²] – [1 × (64a³x³)]

\(\Rightarrow\) [(6x² – 4ax)³] = [216x⁶ – 432 ax⁵ + 288 a²x⁴ – 64a³x³] . . . . . . . . . . (2)

From equation (1) and equation (2)

=[216x⁶ – 432 ax⁵ + 288 a²x⁴ – 64a³x³] + [(36x⁴ + 16a²x² – 48ax³) × 18a²] + [(6x² – 4ax) × 108a⁴] + [ 216a⁶]

=[216x⁶ – 432 ax⁵ + 288 a²x⁴ – 64a³x³ + 648a²x⁴ + 288a⁴x² – 864a³x³ + 648x²a⁴ – 432xa⁵ + 216a⁶]

=8[27x⁶ – 54 ax⁵ + 36 a²x⁴ – 8a³x³ + 81a²x⁴ + 36a⁴x² – 108a³x³ + 81x²a⁴ – 54xa⁵ + 27a⁶]

=8[27x⁶ – 54 ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶]

Therefore, the expansion of (6x² – 4ax + 6a²)³ :

= 8[27x⁶ – 54 ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶]

Q.11: By using the Binomial Theorem, find the expansion of \(\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}\); where ≠ 0

Sol.

The given expression \(\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}\) can be expanded as \(\left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}\)

Now, by using Binomial Theorem:

Therefore, \(\left [\left (1-\frac{5}{x} \right )+\frac{x}{5} \right ]^{4}\)

= [⁴C₀× \(\left (1-\frac{5}{x} \right )^{4}\)] + [⁴C₁× \(\left (1-\frac{5}{x} \right )^{3}\) × \(\left ( \frac{x}{5} \right )\)] + [⁴C₂× \(\left (1-\frac{5}{x} \right )^{2}\) × \(\left ( \frac{x}{5} \right )^{2}\)] + [⁴C₃× \(\left (1-\frac{5}{x} \right )^{1}\) × \(\left ( \frac{x}{5} \right )^{3}\) + [⁴C₄× \(\left ( \frac{x}{5} \right )^{4}\)]

\(\\=\left [ 1\times \left (1-\frac{5}{x} \right )^{4} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{3}\times \left ( \frac{x}{5} \right ) \right ]+\left [ 6\times \left (1-\frac{5}{x} \right )^{2} \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )^{1} \times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ 1 \times \left ( \frac{x}{5} \right )^{4} \right ]\) . . . . . . . . . . . . . . . . . . . (1)

Now, \(\left (1-\frac{5}{x} \right )^{4}:\\\)

By using Binomial Theorem:

Therefore, \(\left (1-\frac{5}{x} \right )^{4}\) = [⁴C₀] – [⁴C₁ × \(\left ( \frac{5}{x} \right )^{1}\)] + [⁴C₂ × \(\left ( \frac{5}{x} \right )^{2}\)] – [⁴C₃ × \(\left ( \frac{5}{x} \right )^{3}\)] + [⁴C₄ × \(\left ( \frac{5}{x} \right )^{4}\)]

\(\\= \left [ 1 \right ] – \left [ 4 \times \left ( \frac{5}{x} \right ) \right ] + \left [ 6 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 4\times \left ( \frac{125}{x^{3}} \right ) \right ] + \left [ 1 \times \left ( \frac{625}{x^{4}} \right ) \right ]\\\)

\(\\=\left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]\) . . . . . . . . (2)

And, \(\left (1-\frac{5}{x} \right )^{3}:\\\)

[ ³C₀] – [ ³C₁ × \(\left ( \frac{5}{x} \right )^{1}\)] + [³C₂ × \(\left ( \frac{5}{x} \right )^{2}\)] – [³C₃ × \(\left ( \frac{5}{x} \right )^{3}\)]

\(\\= [1]-\left [ 3\times \left ( \frac{5}{x} \right ) \right ] + \left [ 3 \times \left ( \frac{25}{x^{2}} \right ) \right ] – \left [ 1\times \left ( \frac{125}{x^{3}} \right ) \right ]\\\)

\(\\=\left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ]\) . . . . . . . . . . . . . . . . . . . (3)

Now, from equation (1), equation (2) and equation (3):

\(\\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ 4 \times \left [ 1-\frac{15}{x}+\frac{75}{x^{2}}-\frac{125}{x^{3}} \right ] \times \left ( \frac{x}{5} \right ) \right ]+\\\\\\+\left [ 6\times \left (1+\frac{25}{x^{2}} -\frac{10}{x}\right ) \times \left ( \frac{x}{5} \right )^{2} \right ]+\left [ 4 \times \left (1-\frac{5}{x} \right )\times \left ( \frac{x}{5} \right )^{3} \right ]+\left [ \left ( \frac{x}{5} \right )^{4} \right ]\\\) \(\\\Rightarrow \left [ 1-\frac{20}{x}+\frac{150}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}} \right ]+\left [ \frac{4x}{5}-12+\frac{60}{x}-\frac{100}{x^{2}} \right ]+\left [ \frac{6x^{2}}{25}+6-\frac{12x}{5}\right ]+\left [ \frac{4x^{3}}{125}-\frac{x^{2}}{25} \right ]+\left [ \frac{x^{4}}{625} \right ]\\\) \(\\\\\Rightarrow \left [\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}-\frac{8x}{5}-5+\frac{40}{x}+ \frac{x^{2}}{5}+\frac{4x^{3}}{125}+\frac{x^{4}}{625} \right ]\\\\\)

Therefore, the expansion of \(\left ( 1+\frac{x}{5}-\frac{5}{x} \right )^{4}\):

= \(\left [\frac{x^{4}}{625}+\frac{4x^{3}}{125}+ \frac{x^{2}}{5}-\frac{8x}{5}-5+\frac{40}{x}+\frac{50}{x^{2}}-\frac{500}{x^{3}}+\frac{625}{x^{4}}\right ]\)

Q.12: By using the Binomial Theorem, find the expansion of (5x³ – 2ax + 3a³)³.

Sol.

The given expression (5x³ – 2ax + 3a³)³ can be expanded as [(5x³ – 2ax) + 3a³)³]

By using Binomial Theorem:

Therefore, [ (5x³ – 2ax) + 3a³)³] = [ ³C₀× (5x³ – 2ax)³] + [ ³C₁× (5x³ – 2ax)² × 3a³] + [ ³C₂× (5x³ – 2ax)¹ × (3a³)²] + [ ³C₃× (5x³ – 2ax)⁰× (3a³)³]

= [1× (5x³ – 2ax)³] + [3 × (5x³ – 2ax)² × 3a³] + [3 × (5x³ – 2ax) × (3a³)²] + [1 × (3a³)³]

= [(5x³ – 2ax)³] + [(5x³ – 2ax)² × 9a³] + [(5x³ – 2ax) × 27a⁶] + [27a⁹] . . . . . . . . . . . . (1)

Now, [(5x³ – 2ax)³]:

By using Binomial Theorem:

Therefore, [(5x² – 2ax)³] = [³C₀× (5x³)³] – [³C₁× (5x³)²) × 2ax] + [³C₂ × 5x³ × (2ax)²] – [³C₃ × (2ax)³]

= [1 × 125x⁹] – [3 × (25x⁶)× 2ax] + [3 × (5x³) × 4a²x²] – [1 × 8a³x³]

[(5x³ – 2ax)³] = [125x⁶ – 150ax⁵ + 60a²x⁴ – 8a³x³] . . . . . . . . . . (2)

From equation (1) and equation (2):

=[125x⁶ – 150ax⁵ + 60a²x⁴ – 8a³x³] + [(25x⁶ + 4a²x² – 20ax⁴) × 9a³] + [(5x³ – 2ax) × 27a⁶] + [ 27a⁹]

=[125x⁶ – 150ax⁵ + 60a²x⁴ – 8a³x³] + [225a³x⁶ + 36a⁵x² – 180a⁴x⁴] + [135a⁶x³ – 54a⁷x ] + [ 27a⁹]

=[125x⁶ – 150ax⁵ + 60a²x⁴ – 8a³x³ + 225a³x⁶ + 36a⁵x² – 180a⁴x⁴ + 135a⁶x³ – 54a⁷x + 27a⁹]

=[125x⁶ + 225a³x⁶ – 150ax⁵ + 60a²x⁴ – 180a⁴x⁴ – 8a³x³ + 135a⁶x³ + 36a⁵x²– 54a⁷x + 27a⁹]

Therefore, the expansion of (5x³ – 2ax + 3a³)³ :

=[125x⁶ + 225a³x⁶ – 150ax⁵ + 60a²x⁴ – 180a⁴x⁴ – 8a³x³ + 135a⁶x³ + 36a⁵x²– 54a⁷x + 27a⁹]

Q.13: Find the coefficient of x⁷ in the product of (1 + x)⁷ (1 + 3x)⁶ by using binomial theorem.

Sol.

By using Binomial Theorem:

Since, (1 + x)ⁿ = [ ⁿC₀ ] + [ ⁿC₁× (x) ] + [ ⁿC₂× (x)²] + [ ⁿC₃× (x³) ] + . . . . . . . . . . . . . . . . . . + [ ⁿC_n× (x)ⁿ]

Therefore, (1 + x)⁷ = [ ⁷C₀] + [ ⁷C₁× (x) ] + [ ⁷C₂× (x)²] + [ ⁷C₃× (x)³] + [ ⁷C₄× (x)⁴ ] + [ ⁷C₅× (x⁵) ] + [ ⁷C₆× (x)⁶ ] + [ ⁷C₇× (x)⁷

\(\\\Rightarrow\) 1 + [ 7x ] + [ 21x² ] + [ 35x³ ] + [ 35x⁴ ] + [ 21x⁵ ] + [ 7x⁶ ] + [ x⁷ ]

Therefore, (1 + x)⁷ = 1+7x + 21x² + 35x³ + 35x⁴ + 21x⁵ + 7x⁶ + x⁷

Now, (1 + 3x)⁶:

By using Binomial Theorem:

(1 + 3x)⁶ = [ ⁶C₀] + [ ⁶C₁× (3x) ] + [ ⁶C₂× (3x)²] + [ ⁶C₃× (3x)³] + [ ⁶C₄× (3x)⁴ ] + [ ⁶C₅× (3x⁵) ] + [ ⁶C₆× (3x)⁶ ]

\(\\\Rightarrow\) 1 + [ 6 × (3x) ] + [ 15 × (9x²) ] + [ 20 × (27x³) ] + [ 15 × (81x⁴) ] + [ 6 × (243x⁵) ] + [ 1 × 729x⁶ ]

\(\\\Rightarrow\) 1 + 18x + 135x² + 540x³ + 1215x⁴ + 1458x⁵+ 729x⁶

Therefore, (1 + 3x)⁶= 1 + 18x + 135x² + 540x³ + 1215x⁴ + 1458x⁵+ 729x⁶

Now, (1 + x)⁷(1 + 3x)⁶:

(1+7x + 21x² + 35x³ + 35x⁴ + 21x⁵ + 7x⁶ + x⁷) × (1 + 18x + 135x² + 540x³ + 1215x⁴ + 1458x⁵+ 729x⁶)

Now, it not required to multiply the above equation, only terms with x⁵ are required and that can be identified by analysing the above equation.

\(\\\Rightarrow\) [{(7x) × (729x⁶)} + {(21x²) × (1458x⁵)} + {(35x³) × (1215x⁴)} + {(35x⁴) × (540x³)} + {(21x⁵) × (135x²)} + {(7x⁶) × (18x)} + {(x⁷) × (1)}]

\(\\\Rightarrow\) [5103x⁵ + 30618x⁵ + 42525x⁵+ 18900x⁵+ 2835x⁷ + 126x⁷+ x⁷]

\(\\\Rightarrow\\\) 72551 x⁷

Therefore, the coefficient of x⁷ = 72551

Q.14: Evaluate \(\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}\)

Sol.

Find (a + b)⁵ – (a – b)⁵

Now, by using Binomial Theorem:

Therefore, [a + b]⁵= [ ⁵C₀× (a⁵) ] + [ ⁵C₁× (a⁴) × b ] + [ ⁵C₂× (a³) × b²] + [ ⁵C₃× (a²) × b³] + [ ⁵C₄× (a¹) × b⁴] + [ ⁵C₅× b⁵]

And, [a – b]⁵ = [ ⁵C₀× (a⁵) ] – [ ⁵C₁× (a⁴) × b ] + [ ⁵C₂× (a³) × b²] – [ ⁵C₃× (a²) × b³] + [ ⁵C₄× (a) × b⁴] – [ ⁵C₅× b⁵]

Therefore, (a + b)⁵– (a – b)⁵ = [⁵C₀ a⁵ + ⁵C₁a⁴b + ⁵C₂a³b²+ ⁵C₃a²b³ + ⁵C₄ab⁴ + ⁵C₅b⁵] – [⁵C₀a⁵ – ⁵C₁a⁴b + ⁵C₂a³b²– ⁵C₃ a²b³+ ⁵C₄ab⁴– ⁵C₅ab⁵]

\(\\\Rightarrow\) [⁵C₀ a⁵ + ⁵C₁a⁴b + ⁵C₂a³b²+ ⁵C₃a²b³ + ⁵C₄ab⁴ + ⁵C₅b⁵ – ⁵C₀a⁵ + ⁵C₁a⁴b – ⁵C₂a³b²+ ⁵C₃ a²b³– ⁵C₄ab⁴+ ⁵C₅ab⁵]

\(\\\Rightarrow\) 2[ ⁵C₁a⁴b + ⁵C₃a²b³+⁵C₅b⁵] = 2[ 5a⁴b + 10a²b³+ b⁵ ]

Therefore, (a + b)⁵– (a – b)⁵ = 2b[ 5a⁴ + 10a²b² + b⁴] . . . . . . (1)

Now, on substituting a = \(\sqrt{3}\) and b = \(\sqrt{7}\) in equation (1) we will get:

\(\\\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}:\\\)

\(\\=2(\sqrt{7})\times \left [ 5( \sqrt{3})^{4} +10\times ( \sqrt{3})^{2}\times (\sqrt{7})^{2}+(\sqrt{7})^{4} \right ]\\\)

\(\\\Rightarrow 2\sqrt{7}\left [\;45+210+49\; \right ]\\\)

Therefore, \(\left(\sqrt{3}+\sqrt{7} \right )^{5}-\left ( \sqrt{3}-\sqrt{7}\right)^{5}\)= \(608\sqrt{7}\)

Q.15: Find the expansion of \(\left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}\)

Sol.

The above given expression can be assumed as: (x + y)⁴ + (x – y)⁴

Now, by using Binomial Theorem:

Therefore, [x + y]⁴= [⁴C₀× (x⁴)] + [⁴C₁× (x³) × (y)] + [⁴C₂× (x²) × (y)²] + [⁴C₃× (x¹) × (y)³] + [⁴C₄× (x⁰) × (y)⁴]

And, [x – y]⁴ = [⁴C₀× (x⁴) ] – [⁴C₁× (x³) × (y) ] + [⁴C₂× (x²) × (y)²] – [⁴C₃× (x¹) × (y)³] + [⁴C₄× (x⁰) × (y)⁴]

Therefore, (x + y)⁴+ (x – y)⁴ = [ ⁴C₀x⁴ + ⁴C₁x³y + ⁴C₂x²y²+ ⁴C₃x y³+ ⁴C₄y⁴] + [ ⁴C₀x⁴ – ⁴C₁x³y + ⁴C₂x²y² – ⁴C₃x y³+ ⁴C₄y⁴]

\(\\\Rightarrow\) [ ⁴C₀x⁴ + ⁴C₁x³y + ⁴C₂x²y²+ ⁴C₃xy³+ ⁴C₄y⁴ + ⁴C₀x⁴ – ⁴C₁x³y + ⁴C₂x²y² – ⁴C₃xy³+ ⁴C₄y⁴]

\(\\\Rightarrow\) 2[ ⁴C₀x⁴ + ⁴C₂x²y² + ⁴C₄y⁴] = 2[ x⁴ + 6x²y² + y⁴]

Therefore, (x + y)⁴+ (x – y)⁴ = 2[ x⁴ + y⁴+ 6 x²y²] . . . . . . . . . . . . . . . . . . . (1)

Now, on substituting x = a² and y = \(\sqrt{a^{2}-5}\) in equation (1) we will get:

\(\\\Rightarrow \left ( a^{2} +\sqrt{a^{2}-5}\right )^{4}+ \left ( a^{2} -\sqrt{a^{2}-5}\right )^{4}:\\\)

\(\\=2\times \left [ (a^{2})^{4}+ \left ( \sqrt{a^{2}-5} \right )^{4}+6\times (a^{2})^{2}\times \left ( \sqrt{a^{2}-5} \right )^{2} \right ]\\\) \(\\\Rightarrow 2\left [ a^{8}+ (a^{2}-5)^{2}+[6a^{4}\times (a^{2}-5)] \right ]\\\) \(\\\Rightarrow 2 \left [ a^{8}+a^{4}+ 25 + 10a^{2}+6a^{6}-30a^{4} \right ]\\\)

Therefore, the expansion of \(\left [ a^{2} +\sqrt{a^{2}-5}\right ]^{4}+ \left [ a^{2} -\sqrt{a^{2}-5}\right ]^{4}:\)

=\(2 \left [ a^{8}+6a^{6}-29a^{4}+10a^{2}+25 \right ]\)