NCERT Solutions Class 11 Maths Chapter 8

Q.1 Using binomial theorem, evaluate the following: (99)⁵

Ans

$\begin{array}{l}\mathrm{Since},{(\mathrm{a}-\mathrm{b})}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}-{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2-...+{(-1)}^{\mathrm{n}}{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ \because {\left(99\right)}^5={(100-1)}^5\\ \mathrm{So},{(100-1)}^5={}^5{\mathrm{C}}_0{\left(100\right)}^5-{}^5{\mathrm{C}}_1{\left(100\right)}^4\left(1\right)+{}^5{\mathrm{C}}_2{\left(100\right)}^3{\left(1\right)}^2\\ -{}^5{\mathrm{C}}_3{\left(100\right)}^2{\left(1\right)}^3+{}^5{\mathrm{C}}_4{\left(100\right)}^1{\left(1\right)}^4-{}^5{\mathrm{C}}_5{\left(1\right)}^5\\ =\left(1\right)\left(10000000000\right)-\left(5\right)\left(100000000\right)\left(1\right)\\ +\left(10\right)\left(1000000\right)\left(1\right)-\left(10\right)\left(10000\right)\left(1\right)\\ +\left(5\right)\left(100\right)\left(1\right)-\left(1\right)\left(1\right)\\ =10000000000-500000000+10000000\\ -100000+500-1\\ =\text{9509900499}\end{array}$

Q.2 Expand the expression: (1 – 2x)⁵.

Ans

$\begin{array}{l}\mathrm{Since},{(1-\mathrm{x})}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0-{}^{\mathrm{n}}{\mathrm{C}}_1\mathrm{x}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{x}}^2-...+{(-1)}^{\mathrm{n}}{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\\ \mathrm{So},{(\text{1}-\text{2x})}^{\text{5}}={}^5{\mathrm{C}}_0-{}^5{\mathrm{C}}_1\left(2\mathrm{x}\right)+{}^5{\mathrm{C}}_2{\left(2\mathrm{x}\right)}^2-{}^5{\mathrm{C}}_3{\left(2\mathrm{x}\right)}^3+{}^5{\mathrm{C}}_4{\left(2\mathrm{x}\right)}^4-{}^5{\mathrm{C}}_5{\left(2\mathrm{x}\right)}^5\\ =1-10\mathrm{x}+40{\mathrm{x}}^2-80{\mathrm{x}}^3+80{\mathrm{x}}^4-32{\mathrm{x}}^5\end{array}$

Q.3

$\mathbf{\text{Expand the expression:}}{\left(\frac{\text{2}}{\text{x}}\text{–}\frac{\text{x}}{\text{2}}\right)}^{\mathbf{\text{5}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\text{Since,\hspace{0.17em}}{\left(\mathrm{a}-\mathrm{b}\right)}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}-{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2-...+{\left(-1\right)}^{\mathrm{n}}{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\frac{2}{\mathrm{x}}-\frac{\mathrm{x}}{2}\right)}^5={}^5{\mathrm{C}}_0{\left(\frac{2}{\mathrm{x}}\right)}^5-{}^5{\mathrm{C}}_1{\left(\frac{2}{\mathrm{x}}\right)}^4\left(\frac{\mathrm{x}}{2}\right)+{}^5{\mathrm{C}}_2{\left(\frac{2}{\mathrm{x}}\right)}^3{\left(\frac{\mathrm{x}}{2}\right)}^2\\ -{}^5{\mathrm{C}}_3{\left(\frac{2}{\mathrm{x}}\right)}^2{\left(\frac{\mathrm{x}}{2}\right)}^3+{}^5{\mathrm{C}}_4\left(\frac{2}{\mathrm{x}}\right){\left(\frac{\mathrm{x}}{2}\right)}^4-{}^5{\mathrm{C}}_5{\left(\frac{\mathrm{x}}{2}\right)}^5\\ =\left(1\right)\left(\frac{32}{{\mathrm{x}}^5}\right)-\left(5\right)\left(\frac{16}{{\mathrm{x}}^4}\right)\left(\frac{\mathrm{x}}{2}\right)+\left(10\right)\left(\frac{8}{{\mathrm{x}}^3}\right)\left(\frac{{\mathrm{x}}^2}{4}\right)\\ -\left(10\right)\left(\frac{4}{{\mathrm{x}}^2}\right)\left(\frac{{\mathrm{x}}^3}{8}\right)+\left(5\right)\left(\frac{2}{\mathrm{x}}\right)\left(\frac{{\mathrm{x}}^4}{16}\right)-\left(1\right)\left(\frac{{\mathrm{x}}^5}{32}\right)\\ =\frac{32}{{\mathrm{x}}^5}-\frac{40}{{\mathrm{x}}^3}+\frac{20}{\mathrm{x}}-5\mathrm{x}+\frac{5}{8}{\mathrm{x}}^3-\frac{{\mathrm{x}}^5}{32}\end{array}$

Q.4 Expand the expression: (2x – 3)⁶.

Ans

$\begin{array}{l}\mathrm{Since},{(\mathrm{a}-\mathrm{b})}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}-{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2-...+{(-1)}^{\mathrm{n}}{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ \mathrm{So},{(\text{2x}-\text{3})}^{\text{6}}={}^6{\mathrm{C}}_0{\left(2\mathrm{x}\right)}^6-{}^6{\mathrm{C}}_1{\left(2\mathrm{x}\right)}^5\left(3\right)+{}^6{\mathrm{C}}_2{\left(2\mathrm{x}\right)}^4{\left(3\right)}^2-{}^6{\mathrm{C}}_3{\left(2\mathrm{x}\right)}^3{\left(3\right)}^3\\ +{}^6{\mathrm{C}}_4{\left(2\mathrm{x}\right)}^2{\left(3\right)}^4-{}^6{\mathrm{C}}_5{\left(2\mathrm{x}\right)}^1{\left(3\right)}^5+{}^6{\mathrm{C}}_6{\left(2\mathrm{x}\right)}^0{\left(3\right)}^6\\ =\left(1\right)2^6{\mathrm{x}}^6-6\times 2^5{\mathrm{x}}^5\times 3+15\times 2^4{\mathrm{x}}^4\times 3^2-20\times 2^3{\mathrm{x}}^3\times 3^3\\ +15\times 2^2{\mathrm{x}}^2\times 3^4-6\times 2^1{\mathrm{x}}^1\times 3^5+\left(1\right)3^6\\ =64{\mathrm{x}}^6-576{\mathrm{x}}^5+2160{\mathrm{x}}^4-4320{\mathrm{x}}^3+4860{\mathrm{x}}^2\\ -2916\mathrm{x}+729\end{array}$

Q.5

$\mathbf{\text{Expand the expression:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\left(\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{x}}\right)}^{\mathbf{\text{5}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{Since},{(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}+{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2+...+{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ \mathrm{So},{(\frac{\mathrm{x}}{3}+\frac{1}{\mathrm{x}})}^5={}^5{\mathrm{C}}_0{\left(\frac{\mathrm{x}}{3}\right)}^5+{}^5{\mathrm{C}}_1{\left(\frac{\mathrm{x}}{3}\right)}^4\left(\frac{1}{\mathrm{x}}\right)+{}^5{\mathrm{C}}_2{\left(\frac{\mathrm{x}}{3}\right)}^3{\left(\frac{1}{\mathrm{x}}\right)}^2\\ +{}^5{\mathrm{C}}_3{\left(\frac{\mathrm{x}}{3}\right)}^2{\left(\frac{1}{\mathrm{x}}\right)}^3+{}^5{\mathrm{C}}_4\left(\frac{\mathrm{x}}{3}\right){\left(\frac{1}{\mathrm{x}}\right)}^4+{}^5{\mathrm{C}}_5{\left(\frac{1}{\mathrm{x}}\right)}^5\\ =\left(1\right)\left(\frac{{\mathrm{x}}^5}{243}\right)+\left(5\right)\left(\frac{{\mathrm{x}}^4}{81}\right)\left(\frac{1}{\mathrm{x}}\right)+\left(10\right)\left(\frac{{\mathrm{x}}^3}{27}\right)\left(\frac{1}{{\mathrm{x}}^2}\right)\\ +\left(10\right)\left(\frac{{\mathrm{x}}^2}{9}\right)\left(\frac{1}{{\mathrm{x}}^3}\right)+\left(5\right)\left(\frac{\mathrm{x}}{3}\right)\left(\frac{1}{{\mathrm{x}}^4}\right)+\left(1\right)\left(\frac{1}{{\mathrm{x}}^5}\right)\\ =\frac{{\mathrm{x}}^5}{243}+\frac{5{\mathrm{x}}^3}{81}+\frac{10\mathrm{x}}{27}+\frac{10}{9\mathrm{x}}+\frac{5}{3{\mathrm{x}}^3}+\frac{1}{{\mathrm{x}}^5}\end{array}$

Q.6

$\mathbf{\text{Expand the expression:}}{\left(\text{x+}\frac{\text{1}}{\text{x}}\right)}^{\mathbf{\text{6}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{Since},{(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}+{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2+...+{}^{\mathrm{n}}{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ \mathrm{So},{(\text{x}+\frac{1}{\mathrm{x}})}^{\text{6}}={}^6{\mathrm{C}}_0{\left(\mathrm{x}\right)}^6+{}^6{\mathrm{C}}_1{\left(\mathrm{x}\right)}^5\left(\frac{1}{\mathrm{x}}\right)+{}^6{\mathrm{C}}_2{\left(\mathrm{x}\right)}^4{\left(\frac{1}{\mathrm{x}}\right)}^2+{}^6{\mathrm{C}}_3{\left(\mathrm{x}\right)}^3{\left(\frac{1}{\mathrm{x}}\right)}^3\\ +{}^6{\mathrm{C}}_4{\left(\mathrm{x}\right)}^2{\left(\frac{1}{\mathrm{x}}\right)}^4+{}^6{\mathrm{C}}_5{\left(\mathrm{x}\right)}^1{\left(\frac{1}{\mathrm{x}}\right)}^5+{}^6{\mathrm{C}}_6{\left(\mathrm{x}\right)}^0{\left(\frac{1}{\mathrm{x}}\right)}^6\\ =\left(1\right){\mathrm{x}}^6+6\times {\mathrm{x}}^4+15\times {\mathrm{x}}^2+20+15\times \frac{1}{{\mathrm{x}}^2}+6\times \frac{1}{{\mathrm{x}}^4}+\left(1\right)\frac{1}{{\mathrm{x}}^6}\\ ={\mathrm{x}}^6+6{\mathrm{x}}^4+15{\mathrm{x}}^2+20+\frac{15}{{\mathrm{x}}^2}+\frac{6}{{\mathrm{x}}^4}+\frac{1}{{\mathrm{x}}^6}\end{array}$

Q.7 Using binomial theorem, evaluate the following: (96)³

Ans

$\begin{array}{l}\mathrm{Since},\\ {\left(\mathrm{a}-\mathrm{b}\right)}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}-{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2-...+{\left(-1\right)}^{\mathrm{n}}{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ \therefore {\left(96\right)}^3={\left(100-4\right)}^3\\ ={}^3{\mathrm{C}}_0{\left(100\right)}^3-{}^3{\mathrm{C}}_1{\left(100\right)}^{3-1}\left(4\right)+{}^3{\mathrm{C}}_2{\left(100\right)}^{3-2}{\left(4\right)}^2\\ -{}^3{\mathrm{C}}_3{\left(100\right)}^{3-3}{\left(4\right)}^3\\ =1000000-3\times 10000\times 4+3\times 100\times 16-1\times 1\times 64\\ =1000000-120000+4800-64\\ =884736\end{array}$

Q.8 Using binomial theorem, evaluate the following: (102)⁵.

Ans

$\begin{array}{l}\mathrm{Since},{(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}+{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2+...+{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ \because {\left(102\right)}^5={(100+2)}^5\\ \mathrm{So},{(100+2)}^5={}^5{\mathrm{C}}_0{\left(100\right)}^5+{}^5{\mathrm{C}}_1{\left(100\right)}^4\left(2\right)+{}^5{\mathrm{C}}_2{\left(100\right)}^3{\left(2\right)}^2\\ +{}^5{\mathrm{C}}_3{\left(100\right)}^2{\left(2\right)}^3+{}^5{\mathrm{C}}_4\left(100\right){\left(2\right)}^4+{}^5{\mathrm{C}}_5{\left(2\right)}^5\\ =\left(1\right)\left(10000000000\right)+\left(5\right)\left(100000000\right)\left(2\right)\\ +\left(10\right)\left(1000000\right)\left(4\right)+\left(10\right)\left(10000\right)\left(8\right)\\ +\left(5\right)\left(100\right)\left(16\right)+\left(1\right)\left(32\right)\\ =10000000000+1000000000+40000000+800000\\ +8000+32\\ =11040808032\end{array}$

Q.9 Using binomial theorem, evaluate the following: (101)⁴

Ans

$\begin{array}{l}\mathrm{Since},{\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}+{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2+...+{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ \because {\left(101\right)}^4={\left(100+2\right)}^4\\ \mathrm{So},{\left(100+1\right)}^4={}^4{\mathrm{C}}_0{\left(100\right)}^4+{}^4{\mathrm{C}}_1{\left(100\right)}^3\left(1\right)+{}^4{\mathrm{C}}_2{\left(100\right)}^2{\left(1\right)}^2\\ +{}^4{\mathrm{C}}_3{\left(100\right)}^1{\left(1\right)}^3+{}^4{\mathrm{C}}_4{\left(1\right)}^4\\ =\left(1\right)\left(100000000\right)+\left(4\right)\left(1000000\right)\left(1\right)\\ +\left(6\right)\left(10000\right)\left(1\right)+\left(4\right)\left(100\right)\left(1\right)+\left(1\right)\left(1\right)\\ =100000000+4000000+60000+400+1\\ =104060401\end{array}$

Q.10 Using Binomial Theorem, indicate which number is larger (1.1)¹⁰⁰⁰⁰ or 1000.

Ans

$\begin{array}{l}{\left(\text{1.1}\right)}^{\text{10000}}\text{=}{\left(\text{1+0.1}\right)}^{\text{10000}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1+10000}\left(\text{0.1}\right)\text{+\hspace{0.17em}higher\hspace{0.17em}powers\hspace{0.17em}}...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=1+1000+\hspace{0.17em}\hspace{0.17em}higher\hspace{0.17em}powers\hspace{0.17em}}...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}>1001\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{Leaving higher power terms}\right)\\ \because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}1001>1000}\\ \text{So,\hspace{0.17em}}{\left(\text{1.1}\right)}^{\text{10000}}\text{>1000.}\end{array}$

Q.11

${\mathbf{\text{Find (a + b)}}}^{\mathbf{\text{4}}}\mathbf{-}{\mathbf{\text{(a}}\mathbf{-}\mathbf{\text{b)}}}^{\mathbf{\text{4}}}\mathbf{\text{. Hence, evaluate}}{\left(\sqrt{\text{3}}\text{+}\sqrt{\text{2}}\right)}^{\mathbf{\text{4}}}\mathbf{-}{(\sqrt{\text{3}}-\sqrt{\text{2}})}^{\mathbf{\text{4}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{Since},{(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}+{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2+...+{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ {\text{(a + b)}}^{\text{4}}-{\text{(a – b)}}^{\text{4}}={(}^4{\mathrm{C}}_0{\mathrm{a}}^4+{}^4{\mathrm{C}}_1{\mathrm{a}}^3\mathrm{b}+{}^4{\mathrm{C}}_2{\mathrm{a}}^2{\mathrm{b}}^2+{}^4{\mathrm{C}}_3{\mathrm{a}}^1{\mathrm{b}}^3+{}^4{\mathrm{C}}_4{\mathrm{b}}^4)\\ -{(}^4{\mathrm{C}}_0{\mathrm{a}}^4-{}^4{\mathrm{C}}_1{\mathrm{a}}^3\mathrm{b}+{}^4{\mathrm{C}}_2{\mathrm{a}}^2{\mathrm{b}}^2-{}^4{\mathrm{C}}_3{\mathrm{a}}^1{\mathrm{b}}^3+{}^4{\mathrm{C}}_4{\mathrm{b}}^4)\\ ={}^4{\mathrm{C}}_0{\mathrm{a}}^4+{}^4{\mathrm{C}}_1{\mathrm{a}}^3\mathrm{b}+{}^4{\mathrm{C}}_2{\mathrm{a}}^2{\mathrm{b}}^2+{}^4{\mathrm{C}}_3{\mathrm{a}}^1{\mathrm{b}}^3+{}^4{\mathrm{C}}_4{\mathrm{b}}^4\\ {-}^4{\mathrm{C}}_0{\mathrm{a}}^4+{}^4{\mathrm{C}}_1{\mathrm{a}}^3\mathrm{b}-{}^4{\mathrm{C}}_2{\mathrm{a}}^2{\mathrm{b}}^2+{}^4{\mathrm{C}}_3{\mathrm{a}}^1{\mathrm{b}}^3-{}^4{\mathrm{C}}_4{\mathrm{b}}^4\\ =2({}^4{\mathrm{C}}_1{\mathrm{a}}^3\mathrm{b}+{}^4{\mathrm{C}}_3{\mathrm{a}}^1{\mathrm{b}}^3)\\ =2(4{\mathrm{a}}^3\mathrm{b}+4{\mathrm{ab}}^3)\\ =8({\mathrm{a}}^3\mathrm{b}+{\mathrm{ab}}^3)\\ \mathrm{Putting}\mathrm{a}=\sqrt{3}\text{}\mathrm{and}\mathrm{b}=\sqrt{2},\mathrm{we}\mathrm{get}\\ {(\sqrt{3}+\sqrt{2})}^4-{(\sqrt{3}-\sqrt{2})}^4\\ =8\{{\left(\sqrt{3}\right)}^3\left(\sqrt{2}\right)+\left(\sqrt{3}\right){\left(\sqrt{2}\right)}^3\}\\ =8\{3\sqrt{6}+2\sqrt{6}\}\\ =8\left(5\sqrt{6}\right)\\ =40\sqrt{6}\end{array}$

Q.12

$\begin{array}{l}{\mathbf{\text{Find (x + 1)}}}^{\mathbf{\text{6}}}\mathbf{\text{+(x}}\mathbf{-}{\mathbf{\text{1)}}}^{\mathbf{\text{6}}}\mathbf{\text{. Hence or otherwise evaluate}}\\ {\left(\sqrt{\text{2}}\text{+1}\right)}^{\mathbf{\text{6}}}\mathbf{-}{(\sqrt{\text{2}}-\text{1})}^{\mathbf{\text{6}}}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},{(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}+{}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}\mathrm{b}+{}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2+...+{}^{\mathrm{n}}{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\mathrm{b}}^{\mathrm{n}}\\ \mathrm{So},{(\mathrm{x}+1)}^{\text{6}}={}^6{\mathrm{C}}_0{\left(\mathrm{x}\right)}^6+{}^6{\mathrm{C}}_1{\left(\mathrm{x}\right)}^5+{}^6{\mathrm{C}}_2{\left(\mathrm{x}\right)}^4+{}^6{\mathrm{C}}_3{\left(\mathrm{x}\right)}^3+{}^6{\mathrm{C}}_4{\left(\mathrm{x}\right)}^2\\ +{}^6{\mathrm{C}}_5{\left(\mathrm{x}\right)}^1+{}^6{\mathrm{C}}_6{\left(\mathrm{x}\right)}^0\\ =\left(1\right){\mathrm{x}}^6+6{\mathrm{x}}^5+15{\mathrm{x}}^4+20{\mathrm{x}}^3+15{\mathrm{x}}^2+6\mathrm{x}+1\\ {(\mathrm{x}-1)}^{\text{6}}={}^6{\mathrm{C}}_0{\left(\mathrm{x}\right)}^6-{}^6{\mathrm{C}}_1{\left(\mathrm{x}\right)}^5+{}^6{\mathrm{C}}_2{\left(\mathrm{x}\right)}^4-{}^6{\mathrm{C}}_3{\left(\mathrm{x}\right)}^3+{}^6{\mathrm{C}}_4{\left(\mathrm{x}\right)}^2\\ -{}^6{\mathrm{C}}_5{\left(\mathrm{x}\right)}^1+{}^6{\mathrm{C}}_6{\left(\mathrm{x}\right)}^0\\ =\left(1\right){\mathrm{x}}^6-6{\mathrm{x}}^5+15{\mathrm{x}}^4-20{\mathrm{x}}^3+15{\mathrm{x}}^2-6\mathrm{x}+1\\ {\text{(x + 1)}}^{\text{6}}{\text{+(x – 1)}}^{\text{6}}\\ =2{\mathrm{x}}^6+30{\mathrm{x}}^4+30{\mathrm{x}}^2+2\\ =2({\mathrm{x}}^6+15{\mathrm{x}}^4+15{\mathrm{x}}^2+1)\\ \mathrm{Putting}\text{x}=\sqrt{2},\text{we get}\\ {(\sqrt{2}+1)}^6-{(\sqrt{2}-1)}^6\\ =2\{{\left(\sqrt{2}\right)}^6+15{\left(\sqrt{2}\right)}^4+15{\left(\sqrt{2}\right)}^2+1\}\\ =2(8+60+30+1)\\ =2\left(99\right)\\ =198\end{array}$

Q.13 Show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is a positive integer.

Ans

$\begin{array}{l}{9}^{\text{n}+1}={\left(1+8\right)}^{\text{n}+1}\\ =1+\left(\text{n}+1\right)8+\frac{\left(\text{n}+1\right)\text{n}}{2!}{8}^2+\frac{\left(\text{n}+1\right)\text{n}\left(\text{n}-1\right)}{3!}{8}^3+...\\ \Rightarrow 9^{\text{n}+1}-1-8\text{n}-8=\frac{\left(\text{n}+1\right)\text{n}}{2!}{8}^2+\frac{\left(\text{n}+1\right)\text{n}\left(\text{n}-1\right)}{3!}{8}^3+...\\ 9^{\text{n}+1}-8\text{n}-9=8^2\left\{\frac{\left(\text{n}+1\right)\text{n}}{2!}+\frac{\left(\text{n}+1\right)\text{n}\left(\text{n}-1\right)}{3!}8+...\right\}\\ 9^{\text{n}+1}-8\text{n}-9=64\left\{\frac{\left(\text{n}+1\right)\text{n}}{2!}+\frac{\left(\text{n}+1\right)\text{n}\left(\text{n}-1\right)}{3!}8+...\right\}\\ 9^{\text{n}+1}-8\text{n}-9=\mathrm{multiple}\text{of 64, because n is an integer.}\\ \Rightarrow 9^{\text{n}+1}-8\text{n}-9\text{is divisible by 64.}\end{array}$

Q.14

$\mathbf{\text{Prove that}}\sum _{\text{r=0}}^{\text{n}}{\text{3}}^{\text{r}}{}^{\text{n}}{\text{C}}_{\text{r}}{\text{\hspace{0.17em}=4}}^{\text{n}}\text{.}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\sum _{\mathrm{r}=0}^{\mathrm{n}}{3}^{\mathrm{r}}{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=3^0{}^{\mathrm{n}}{\mathrm{C}}_0+3^1{}^{\mathrm{n}}{\mathrm{C}}_1+3^2{}^{\mathrm{n}}{\mathrm{C}}_2+3^3{}^{\mathrm{n}}{\mathrm{C}}_3+3^4{}^{\mathrm{n}}{\mathrm{C}}_4+...\\ ={}^{\mathrm{n}}{\mathrm{C}}_0.3^0{+}^{\mathrm{n}}{\mathrm{C}}_1.3^1{+}^{\mathrm{n}}{\mathrm{C}}_2.3^2{+}^{\mathrm{n}}{\mathrm{C}}_3.3^3+...\\ ={(1+3)}^{\mathrm{n}}\left[\begin{array}{l}\because {(1+\mathrm{x})}^{\mathrm{n}}={}^{\mathrm{n}}{\mathrm{C}}_0.{\mathrm{x}}^0+{}^{\mathrm{n}}{\mathrm{C}}_1.{\mathrm{x}}^1+{}^{\mathrm{n}}{\mathrm{C}}_2.{\mathrm{x}}^2\\ +{}^{\mathrm{n}}{\mathrm{C}}_3.{\mathrm{x}}^3+...\end{array}\right]\\ =4^{\mathrm{n}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.15 Find the coefficient of x⁵ in (x + 3)⁸.

Ans

$\begin{array}{l}{\text{The (r + 1)}}^{\text{th}}{\text{term of the expansion (x + y)}}^{\text{n}}\text{is given by}\\ {\text{T}}_{\text{r\hspace{0.17em}+1}}={}^{\text{n}}{\text{C}}_{\text{r}}{{\text{x}}^{\text{n}}}^{-\text{r}}{\text{y}}^{\text{r}}\text{.}\\ \text{In the expansion of}{(\text{x}+\text{3})}^{\text{8}},\\ {\text{T}}_{\text{r\hspace{0.17em}+1}}={}^{\text{8}}{\text{C}}_{\text{r}}{{\text{x}}^{\text{8}}}^{-\text{r}}{\text{\hspace{0.17em}3}}^{\text{r}}\\ \mathrm{For}{\text{the coefficient of x}}^{\text{5}},\\ \text{let \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}8}-\mathrm{r}=5\\ \Rightarrow \mathrm{r}=8-5=3\\ \mathrm{So},{\text{the coefficient of x}}^{\text{5}}\\ ={}^{\text{8}}{\text{C}}_{\text{3}}{\text{\hspace{0.17em}3}}^{\text{3}}\\ =27\times \frac{8!}{3!5!}\\ =27\times \frac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}\\ =1512\end{array}$

Q.16 Find the coefficient of a⁵b⁷ in (a – 2b)¹².

Ans

$\begin{array}{l}{\text{The (r + 1)}}^{\text{th}}{\text{term of the expansion (x + y)}}^{\text{n}}\text{is given by}\\ {\text{T}}_{\text{r\hspace{0.17em}+1}}={}^{\text{n}}{\text{C}}_{\text{r}}{{\text{x}}^{\text{n}}}^{-\text{r}}{\text{y}}^{\text{r}}\text{.}\\ \text{In the expansion of}{(\text{a}-\text{2b})}^{\text{12}},\\ {\text{T}}_{\text{r\hspace{0.17em}+1}}={}^{\text{12}}{\text{C}}_{\text{r}}{{\text{a}}^{\text{12}}}^{-\text{r}}{(-\text{2b})}^{\text{r}}\\ \mathrm{For}{\text{the coefficient of a}}^{\text{5}}{\text{b}}^{\text{7}},\\ \text{let 12}-\mathrm{r}=5\\ \Rightarrow \mathrm{r}=12-5\\ \begin{array}{l}=7\\ \mathrm{So},{\text{the coefficient of a}}^{\text{5}}{\text{b}}^{\text{7}}\\ ={}^{\text{12}}{\text{C}}_{\text{7}}{(-2)}^{\text{7}}\\ =-128\times \frac{12!}{7!5!}\\ =-128\times \frac{12\times 11\times 10\times 9\times 8\times 7!}{7!\times 5\times 4\times 3\times 2\times 1}\\ =-101376\end{array}\end{array}$

Q.17 Write the general term in the expansion of (x² – y)⁶.

Ans

$\begin{array}{l}\mathrm{The}\text{general term of}{({\text{x}}^{\text{2}}–\text{y})}^{\text{6}}\mathrm{is}\\ {\mathrm{T}}_{\mathrm{r}+1}={}^6{\mathrm{C}}_{\mathrm{r}}{\left({\mathrm{x}}^2\right)}^{6-\mathrm{r}}{(-\mathrm{y})}^{\mathrm{r}}\\ ={(-1)}^{\mathrm{r}}{}^6{\mathrm{C}}_{\mathrm{r}}{\left(\mathrm{x}\right)}^{12-2\mathrm{r}}.{\mathrm{y}}^{\mathrm{r}}\end{array}$

Q.18 Find the 4^th term in the expansion of (x – 2y)¹².

Ans

$\begin{array}{l}\mathrm{Let}{\text{ 4}}^{\text{th}}\text{term in the expression of}{(\text{x}-\text{2y})}^{\text{12}}{\text{be T}}_{\text{4}}\text{,\hspace{0.17em}then}\\ {\text{T}}_{\text{4}}={\mathrm{T}}_{3+1}\\ ={}^{12}{\mathrm{C}}_3{\mathrm{x}}^{12-3}{(-2\mathrm{y})}^3\\ ={}^{12}{\mathrm{C}}_3{\mathrm{x}}^9{(-2)}^3{\mathrm{y}}^3\\ =-8\times \frac{12!}{3!9!}{\mathrm{x}}^9{\mathrm{y}}^3\\ =-8\times \frac{12\times 11\times 10\times 9!}{3\times 2\times 1\times 9!}{\mathrm{x}}^9{\mathrm{y}}^3\\ =-1760{\mathrm{x}}^9{\mathrm{y}}^3\end{array}$

Q.19

${\mathbf{\text{Find the 13}}}^{\mathbf{\text{th}}}\mathbf{\text{term in the expansion of\hspace{0.17em}\hspace{0.17em}}}{(\text{9x}-\frac{\text{1}}{\text{3}\sqrt{\text{x}}})}^{\mathbf{\text{18}}}\mathbf{\text{,\hspace{0.17em}x}}\mathbf{\ne }\mathbf{\text{0.}}$

Ans

$\begin{array}{l}{\text{Let 13}}^{\text{th}}\text{term in the expression of}{\left(9\mathrm{x}-\frac{1}{3\sqrt{\mathrm{x}}}\right)}^{18}{\text{be T}}_{\text{13}}\text{,\hspace{0.17em}then}\\ {\text{T}}_{\text{13}}={\mathrm{T}}_{12+1}\\ ={}^{18}{\mathrm{C}}_{12}{\left(9\mathrm{x}\right)}^{18-12}{\left(-\frac{1}{3\sqrt{\mathrm{x}}}\right)}^{12}\\ ={}^{18}{\mathrm{C}}_{12}\left(9^6{\mathrm{x}}^6\right){\left(-\frac{1}{3}\right)}^{12}\left(\frac{1}{{\mathrm{x}}^6}\right)\\ =\frac{9^6}{3^{12}}\times \frac{18!}{12!6!}{\mathrm{x}}^6\left(\frac{1}{{\mathrm{x}}^6}\right)\\ =\frac{18\times 17\times 16\times 15\times 14\times 13}{6\times 5\times 4\times 3\times 2}\times \frac{{\mathrm{x}}^6}{{\mathrm{x}}^6}\\ =18564\end{array}$

Q.20

$\mathbf{Write}{\mathbf{\text{the general term in the expansion of (x}}}^{\mathbf{2}}\mathbf{-}\mathbf{yx}{\mathbf{\text{)}}}^{\mathbf{12}}\mathbf{,}\mathbf{x}\mathbf{\ne }\mathbf{0}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{The}\text{general term of}{({\text{x}}^{\text{2}}-\text{yx})}^{\text{12}}\mathrm{is}\\ {\mathrm{T}}_{\mathrm{r}+1}={}^{12}{\mathrm{C}}_{\mathrm{r}}{\left({\mathrm{x}}^2\right)}^{12-\mathrm{r}}{(-\mathrm{yx})}^{\mathrm{r}}\\ ={(-1)}^{\mathrm{r}}{}^{12}{\mathrm{C}}_{\mathrm{r}}{\left(\mathrm{x}\right)}^{24-2\mathrm{r}}.{\mathrm{x}}^{\mathrm{r}}.{\mathrm{y}}^{\mathrm{r}}\\ ={(-1)}^{\mathrm{r}}{}^{12}{\mathrm{C}}_{\mathrm{r}}{\left(\mathrm{x}\right)}^{24-\mathrm{r}}.{\mathrm{y}}^{\mathrm{r}}\end{array}$

Q.21

$\mathbf{\text{Find the middle term in the expansion of\hspace{0.17em}\hspace{0.17em}}}{(\text{3}-\frac{{\text{x}}^{\text{3}}}{\text{6}})}^{\mathbf{\text{7}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}{(3-\frac{{\mathrm{x}}^3}{6})}^7\\ \mathrm{Here},\mathrm{n}=7\left(\mathrm{odd}\right)\\ \mathrm{So},\text{middle terms in expansion}\\ ={\left(\frac{7+1}{2}\right)}^{\mathrm{th}}\mathrm{term}\text{and}{(\frac{7+1}{2}+1)}^{\mathrm{th}}\mathrm{term}\\ =4^{\mathrm{th}}{\text{ term and 5}}^{\text{th}}\text{term}\\ {\text{Then,\hspace{0.17em}\hspace{0.17em}T}}_{\text{4}}={\mathrm{T}}_{3+1}\\ ={}^7{\mathrm{C}}_3.3^{7-3}.{(-\frac{{\mathrm{x}}^3}{6})}^3\\ ={}^7{\mathrm{C}}_3.3^4.{(-\frac{1}{6})}^3{\mathrm{x}}^9\\ =-\frac{35\times 27\times 3}{36\times 6}{\mathrm{x}}^9\\ =-\frac{105}{8}{\mathrm{x}}^9\\ \mathrm{and}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{5}}={\mathrm{T}}_{4+1}\end{array}$ $\begin{array}{l}={}^7{\mathrm{C}}_4.3^{7-4}.{(-\frac{{\mathrm{x}}^3}{6})}^4\\ ={}^7{\mathrm{C}}_4.3^3{(-\frac{1}{6})}^4{\mathrm{x}}^{12}\\ =\frac{35\times 27}{36\times 6\times 6}{\mathrm{x}}^{12}\\ =\frac{35}{48}{\mathrm{x}}^{12}\\ \mathrm{Thus},\text{the middle terms of the expansion are}-\frac{105}{8}{\text{x}}^{\text{9}}\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\frac{35}{48}{\mathrm{x}}^{12}.\end{array}$

Q.22

$\mathbf{\text{Find the middle term in the expansion of\hspace{0.17em}\hspace{0.17em}}}{\left(\frac{\text{x}}{\text{3}}\text{+9y}\right)}^{\mathbf{\text{10}}}\mathbf{\text{.}}$

Ans

$\begin{array}{l}\mathrm{Given}:{\left(\frac{\text{x}}{3}+9\text{y}\right)}^{10}\\ \mathrm{Here},\text{\hspace{0.33em}n}=10\left(\mathrm{even}\right)\\ \mathrm{The}\mathrm{middle}\mathrm{term}\mathrm{in}\mathrm{the}\text{expansion}={\left(\frac{10}{2}+1\right)}^{\mathrm{th}}\text{term}\\ ={\text{6}}^{\text{th}}\text{\hspace{0.33em}term}\\ {\text{Then,\hspace{0.33em}T}}_{\text{6}}={\text{T}}_{5+1}\\ ={}^{10}{\text{C}}_5{\left(\frac{\text{x}}{3}\right)}^{10-5}{\left(9\text{y}\right)}^5\\ =\frac{10!}{5!\left(10-5\right)!}\times \frac{{\text{x}}^5}{3^5}\times 9^5{\text{y}}^5\\ =\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}\times \frac{{\text{x}}^5}{3^5}\times 3^{10}{\text{y}}^5\\ =6\times 42\times 9\times 27{\text{x}}^5{\text{y}}^4\\ =61236{\text{\hspace{0.33em}x}}^5{\text{y}}^5\\ \mathrm{Thus},\text{the middle term of given expression is}61236{\text{\hspace{0.33em}x}}^5{\text{y}}^5.\end{array}$

Q.23 In the expansion of (1 + a)^m+n, prove that coefficients of a^m and aⁿ are equal.

Ans

$\begin{array}{l}\mathrm{The}\text{given expression is}{(\text{1}+\text{a})}^{\text{m}+\text{n}}.\\ \mathrm{The}\text{general term}\left({\mathrm{T}}_{\mathrm{r}+1}\right)={}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{a}}^{\mathrm{r}}...\left(\mathrm{i}\right)\\ \mathrm{Putting}\text{r}=\mathrm{m}\text{in equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{Coefficient}{\text{of a}}^{\text{m}}={}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{m}}\\ \mathrm{Putting}\text{r}=\mathrm{n}\text{in equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{Coefficient}{\text{of a}}^{\text{n}}={}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\\ ={}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{m}+\mathrm{n}-\mathrm{n}}[\because {}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}]\\ ={}^{\mathrm{m}+\mathrm{n}}{\mathrm{C}}_{\mathrm{m}}\\ \mathrm{Thus},\\ {\text{the coefficient of a}}^{\text{m}}={\text{the coefficient of a}}^{\text{n}}.\end{array}$

Q.24 The coefficients of the (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1 : 3 : 5. Find n and r.

Ans

$\begin{array}{l}\text{The coefficients of the}{(\text{r}-\text{1})}^{\text{th}},{\text{r}}^{\text{th}}\text{and}\mathrm{}{(\text{r}+\text{1})}^{\text{th}}\text{terms in the}\\ \text{expansion of}{(\text{x}+\text{1})}^{\text{n}}\mathrm{}\text{are in the ratio 1}:\text{3}:\text{5.Then,}\\ {\text{\hspace{0.17em}T}}_{\text{r}-\text{1}}={\text{T}}_{(\text{r}-\text{2})\text{\hspace{0.17em}+1}}\\ ={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-2}{\mathrm{x}}^{\mathrm{n}-(\mathrm{r}-2)}\\ =\frac{\mathrm{n}!}{(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+2)!}{\mathrm{x}}^{(\mathrm{n}-\mathrm{r}+2)}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{r}}={\text{T}}_{(\text{r}-\text{1})\text{\hspace{0.17em}+1}}\\ ={}^{\mathrm{n}}{\mathrm{C}}_{(\mathrm{r}-1)}{\mathrm{x}}^{\mathrm{n}-(\mathrm{r}-1)}\\ =\frac{\mathrm{n}!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r}+1)!}{\mathrm{x}}^{(\mathrm{n}-\mathrm{r}+1)}\\ {\text{\hspace{0.17em}\hspace{0.17em}T}}_{\text{r}+\text{1}}={\text{T}}_{\mathrm{r}}\\ ={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{n}-\mathrm{r}}\\ =\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}{\mathrm{x}}^{(\mathrm{n}-\mathrm{r})}\\ \mathrm{Since},\\ \frac{\mathrm{Coefficient}\text{of}{\mathrm{T}}_{\mathrm{r}-1}}{\mathrm{Coefficient}\text{of}{\mathrm{T}}_{\mathrm{r}}}=\frac{\frac{\mathrm{n}!}{(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+2)!}}{\frac{\mathrm{n}!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r}+1)!}}\\ \frac{1}{3}=\frac{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r}+1)!}{(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+2)!}\\ \frac{1}{3}=\frac{(\mathrm{r}-1).(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+1)!}{(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+2).(\mathrm{n}-\mathrm{r}+1)!}\end{array}$ $\begin{array}{l}\frac{1}{3}=\frac{(\mathrm{r}-1)}{(\mathrm{n}-\mathrm{r}+2)}\\ \mathrm{n}-\mathrm{r}+2=3\mathrm{r}-3\\ \mathrm{n}-4\mathrm{r}=-5...\left(\mathrm{i}\right)\\ \frac{\mathrm{Coefficient}\text{of}{\mathrm{T}}_{\mathrm{r}}}{\mathrm{Coefficient}\text{of}{\mathrm{T}}_{\mathrm{r}+1}}=\frac{\frac{\mathrm{n}!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r}+1)!}}{\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}}\\ \frac{3}{5}=\frac{\mathrm{r}.(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r})!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r}+1).(\mathrm{n}-\mathrm{r})!}\\ \frac{3}{5}=\frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}\\ 3\mathrm{n}-3\mathrm{r}+3=5\mathrm{r}\\ 3\mathrm{n}-8\mathrm{r}=-3...\left(\mathrm{ii}\right)\\ \mathrm{On}\mathrm{solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{n}=\text{7 and r}=\text{3.}\end{array}$

Q.25 Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n – 1}.

Ans

$\begin{array}{l}\mathrm{Given}\text{expressions are:\hspace{0.17em}\hspace{0.17em}}{(\text{1}+\text{x})}^{\text{2n}}\text{and}{(\text{1}+\text{x})}^{\text{2n-1}}\\ \mathrm{For}{(\text{1}+\text{x})}^{\text{2n}}:\\ \mathrm{Since},{\text{T}}_{\text{r+1}}={}^{2\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}\\ \mathrm{Putting}\text{r}=\text{n, we get}\\ {\text{coefficient of x}}^{\text{n}}={}^{2\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}\end{array}$ $\begin{array}{l}=\frac{2\mathrm{n}!}{\mathrm{n}!(2\mathrm{n}-\mathrm{n})!}\\ =\frac{2\mathrm{n}!}{\mathrm{n}!\mathrm{n}!}\\ =\frac{2\mathrm{n}(2\mathrm{n}-1)!}{\mathrm{n}(\mathrm{n}-1)!\mathrm{n}!}\\ =2\times \frac{(2\mathrm{n}-1)!}{(\mathrm{n}-1)!\mathrm{n}!}\\ \mathrm{For}{(\text{1}+\text{x})}^{\text{2n-1}}:\\ \mathrm{Since},{\mathrm{T}}_{\mathrm{r}+1}={}^{2\mathrm{n}-1}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}\\ \mathrm{Putting}\text{r}=\text{n, we get}\\ {\text{coefficient of x}}^{\text{n}}={}^{2\mathrm{n}-1}{\mathrm{C}}_{\mathrm{n}}\\ =\frac{(2\mathrm{n}-1)!}{\mathrm{n}!(2\mathrm{n}-1-\mathrm{n})!}\\ =\frac{(2\mathrm{n}-1)!}{\mathrm{n}!(\mathrm{n}-1)!}\\ \mathrm{Thus},{\text{coefficient of x}}^{\text{n}}\text{in}{(1+\mathrm{x})}^{\mathrm{n}}\text{is double of the coefficient}\\ {\text{of x}}^{\text{n}}\text{in}{(1+\mathrm{x})}^{2\mathrm{n}-1}.\end{array}$

Q.26 Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.

Ans

$\begin{array}{l}{\text{Given: Coefficient of x}}^{\text{2}}\text{in the expansion of}{\left(\text{1}+\text{x}\right)}^{\text{m}}\mathrm{is}6.\\ {\text{The (r + 1)}}^{\text{th}}\text{term of the expansion}{\left(\text{1}+\text{x}\right)}^{\text{m}}\text{is given by}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{r+1}}={}^{\text{m}}{\text{C}}_{\text{r}}{\text{x}}^{\text{r}}\\ \text{Putting r}=\text{2, we get}\\ {\text{coefficient of x}}^{\text{2}}={}^{\text{m}}{\text{C}}_{\text{2}}\\ 6=\frac{\mathrm{m}!}{2!\left(\mathrm{m}-2\right)!}\\ \Rightarrow 6=\frac{\mathrm{m}\left(\mathrm{m}-1\right)\left(\mathrm{m}-2\right)!}{2\left(\mathrm{m}-2\right)!}\\ \Rightarrow 12=\mathrm{m}\left(\mathrm{m}-1\right)\\ \Rightarrow 4\times 3=\mathrm{m}\left(\mathrm{m}-1\right)\\ \Rightarrow 4\times \left(4-1\right)=\mathrm{m}\left(\mathrm{m}-1\right)\\ \Rightarrow \mathrm{m}=4\end{array}$

Q.27 Find a, b and n in the expansion of (a + b)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Ans

$\begin{array}{l}\mathrm{Three}\text{terms in the expansion of}{(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}\text{are\hspace{0.17em}729, 7290 and}\\ \text{30375 respectively.}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}T}}_{\text{1}}=\text{729\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \Rightarrow {\mathrm{T}}_{0+1}={}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}{\mathrm{b}}^0\\ \Rightarrow {}^{\mathrm{n}}{\mathrm{C}}_0{\mathrm{a}}^{\mathrm{n}}=\text{729}\\ \Rightarrow {\mathrm{a}}^{\mathrm{n}}=729...\left(\mathrm{i}\right)\\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}T}}_{\text{2}}=\text{729}0\\ \Rightarrow {\mathrm{T}}_{1+1}={}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{a}}^{\mathrm{n}-1}{\mathrm{b}}^1\\ \Rightarrow {\mathrm{na}}^{\mathrm{n}-1}\mathrm{b}=\text{729}0...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}\Rightarrow \frac{{\mathrm{na}}^{\mathrm{n}}\mathrm{b}}{\mathrm{a}}=\text{729}0\\ \Rightarrow \frac{\mathrm{n}\left(729\right)\mathrm{b}}{\mathrm{a}}=\text{729}0\left[\mathrm{From}\text{eqution}\left(\mathrm{i}\right)\right]\\ \Rightarrow \frac{\mathrm{nb}}{\mathrm{a}}=\frac{7290}{729}\\ \Rightarrow \frac{\mathrm{nb}}{\mathrm{a}}=10...\left(\mathrm{iii}\right)\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}T}}_{\text{3}}=\text{30375}\\ \Rightarrow {\mathrm{T}}_{2+1}=\text{30375}\\ \Rightarrow {}^{\mathrm{n}}{\mathrm{C}}_2{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2=\text{30375}\\ \Rightarrow \frac{\mathrm{n}!}{2!(\mathrm{n}-2)!}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2=\text{30375}\\ \Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2=\text{30375}...\left(\mathrm{iv}\right)\\ \mathrm{Dividing}\text{equation}\left(\mathrm{iv}\right)\text{by equation}\left(\mathrm{ii}\right),\text{we get}\\ \frac{\frac{\mathrm{n}(\mathrm{n}-1)}{2}{\mathrm{a}}^{\mathrm{n}-2}{\mathrm{b}}^2}{{\mathrm{na}}^{\mathrm{n}-1}\mathrm{b}}=\frac{30375}{7290}\\ \Rightarrow \frac{(\mathrm{n}-1)\mathrm{b}}{\mathrm{a}}=\frac{2\times 30375}{7290}\\ \Rightarrow \frac{\mathrm{nb}}{\mathrm{a}}-\frac{\mathrm{b}}{\mathrm{a}}=\frac{25}{3}\\ \Rightarrow 10-\frac{\mathrm{b}}{\mathrm{a}}=\frac{25}{3}\left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\right]\\ \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=10-\frac{25}{3}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{5}{3}\\ \mathrm{From}\text{equation}\left(\mathrm{iii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n}\left(\frac{5}{3}\right)=10\\ \Rightarrow \mathrm{n}=10\times \frac{3}{5}\\ =6\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right),\text{we have}\\ {\text{a}}^{\text{6}}=729\\ {\text{a}}^{\text{6}}=3^6\\ \Rightarrow \mathrm{a}=3\\ \mathrm{Putting}\text{the values of n and a in equation}\left(\mathrm{iii}\right),\text{we get}\\ \frac{6\mathrm{b}}{3}=10\\ \Rightarrow \mathrm{b}=\frac{10}{2}\\ =5\\ \mathrm{Thus},\text{a}=3,\text{b}=5\text{and n}=\text{6.}\end{array}$