NCERT Solutions Class 11 Maths Chapter 4

Q.1 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{+}\mathbf{3}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{{\mathbf{3}}^{\mathbf{n}}}^{\mathbf{–}\mathbf{1}}\mathbf{=}\frac{\mathbf{(}{\mathbf{3}}^{\mathbf{n}}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{2}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):1+3+3^2+...+{3^{\mathrm{n}}}^{–1}=\frac{(3^{\mathrm{n}}-1)}{2}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{3^1-1}{2}=1=\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1+3+3^2+...+{3^{\mathrm{k}}}^{–1}=\frac{(3^{\mathrm{k}}-1)}{2}...\left(1\right)\\ \text{We need to prove that P(k + 1) is true whenever P(k) is true.}\\ \text{We have}\\ 1+3+3^2+...+{3^{\mathrm{k}}}^{–1}+3^{\mathrm{k}}=\frac{(3^{\mathrm{k}+1}-1)}{2}\\ \Rightarrow \frac{(3^{\mathrm{k}}-1)}{2}+3^{\mathrm{k}}=\frac{(3^{\mathrm{k}+1}-1)}{2}...\left(2\right)\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{(3^{\mathrm{k}}-1)}{2}+3^{\mathrm{k}}=\frac{(3^{\mathrm{k}}-1+2{.3}^{\mathrm{k}})}{2}\\ =\frac{3{.3}^{\mathrm{k}}-1}{2}\\ =\frac{3^{\mathrm{k}+1}-1}{2}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.2 Prove by using principal of mathematical induction for all n ∈ N

${\mathbf{1}}^{\mathbf{3}}\mathbf{+}{\mathbf{2}}^{\mathbf{3}}\mathbf{+}{\mathbf{3}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathbf{n}}^{\mathbf{3}}\mathbf{=}{\mathbf{\left\{}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}}\mathbf{\right\}}}^{\mathbf{2}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):1^3+2^3+3^3+...+{\mathrm{n}}^3={\left\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right\}}^2\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1^3=1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:{\left\{\frac{1.(1+1)}{2}\right\}}^2=1.\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1^3+2^3+3^3+...++{\mathrm{n}}^{\mathrm{k}}={\left\{\frac{\mathrm{k}(\mathrm{k}+1)}{2}\right\}}^2...\left(\mathrm{i}\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1^3+2^3+3^3+...++{\mathrm{k}}^3+{(\mathrm{k}+1)}^3={\left\{\frac{(\mathrm{k}+1)(\mathrm{k}+2)}{2}\right\}}^2\\ \Rightarrow {\left\{\frac{\mathrm{k}(\mathrm{k}+1)}{2}\right\}}^2+{(\mathrm{k}+1)}^3={\left\{\frac{(\mathrm{k}+1)(\mathrm{k}+2)}{2}\right\}}^2...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:{\left\{\frac{\mathrm{k}(\mathrm{k}+1)}{2}\right\}}^2+{(\mathrm{k}+1)}^3={(\mathrm{k}+1)}^2\{\frac{\mathrm{k}}{4}+\mathrm{k}+1\}\\ ={(\mathrm{k}+1)}^2\left\{\frac{{\mathrm{k}}^2+4\mathrm{k}+4}{4}\right\}\\ ={(\mathrm{k}+1)}^2\left\{\frac{{(\mathrm{k}+2)}^2}{4}\right\}\\ ={\left\{\frac{(\mathrm{k}+1)(\mathrm{k}+2)}{2}\right\}}^2=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by principle}\\ \text{of mathematical induction, P(n) is true for every positive integer n.}\end{array}$

Q.3 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{)}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{3}\mathbf{)}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{3}\mathbf{+}\mathbf{4}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\mathbf{n}\mathbf{)}}\mathbf{=}\frac{\mathbf{2}\mathbf{n}}{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+4+...+\mathrm{n})}=\frac{2\mathrm{n}}{\mathrm{n}+1}\\ \mathrm{For}\text{n}=1,\\ \Rightarrow \mathrm{P}\left(1\right)=1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{2\left(1\right)}{1+1}=1.\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+4+...+\mathrm{k})}\\ =\frac{2\mathrm{k}}{\mathrm{k}+1}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+4+...+\mathrm{k})}+\\ \frac{1}{\{1+2+3+4+...+(\mathrm{k}+1)\}}=\frac{2(\mathrm{k}+1)}{\mathrm{k}+2}\\ \Rightarrow \frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{1}{\{1+2+3+4+...+(\mathrm{k}+1)\}}=\frac{2(\mathrm{k}+1)}{\mathrm{k}+2}...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{1}{\{1+2+3+4+...+(\mathrm{k}+1)\}}=\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{1}{\left\{\frac{(\mathrm{k}+1)(\mathrm{k}+1+1)}{2}\right\}}\\ =\frac{2\mathrm{k}}{\mathrm{k}+1}+\frac{2}{(\mathrm{k}+1)(\mathrm{k}+2)}\\ =\frac{2}{\mathrm{k}+1}(\mathrm{k}+\frac{2}{\mathrm{k}+2})\\ =\frac{2}{\mathrm{k}+1}\left(\frac{{\mathrm{k}}^2+2\mathrm{k}+1}{\mathrm{k}+2}\right)\\ =\frac{2}{\mathrm{k}+1}\times \frac{{(\mathrm{k}+1)}^2}{\mathrm{k}+2}\\ =\frac{2(\mathrm{k}+1)}{(\mathrm{k}+1)+1}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.4 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{)}}{\mathbf{4}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):1.2.3+2.3.4+...+\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(\mathrm{n}+3)}{4}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1.2.3=6\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{1(1+1)(1+2)(1+3)}{4}=\frac{1\left(2\right)\left(3\right)\left(4\right)}{4}\\ =6\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.2.3+2.3.4+...+\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)\\ =\frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}{4}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.2.3+2.3.4+...+\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)+(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)\\ =\frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)(\mathrm{k}+4)}{4}\\ \Rightarrow \frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}{4}+(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)\\ =\frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)(\mathrm{k}+4)}{4}...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}{4}+(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)\\ =(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)(\frac{\mathrm{k}}{4}+1)\\ =\frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)(\mathrm{k}+4)}{4}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.5 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{.3}\mathbf{+}\mathbf{2}{\mathbf{.3}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{.3}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{..}{\mathbf{\text{n.3}}}^{\mathbf{\text{n}}}\mathbf{=}\frac{(2\text{n}–1){\mathbf{3}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{+}\mathbf{3}}{\mathbf{4}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):1.3+2{.3}^2+3{.3}^3+...+\mathrm{n}.3^{\mathrm{n}}=\frac{(2\mathrm{n}-1)3^{\mathrm{n}+1}+3}{4}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1.3=3\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{(2\times 1-1)3^{1+1}+3}{4}=\frac{12}{4}\\ =3\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.3+2{.3}^2+3{.3}^3+...+\mathrm{k}.3^{\mathrm{k}}=\frac{(2\mathrm{k}-1)3^{\mathrm{k}+1}+3}{4}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.3+2{.3}^2+3{.3}^3+...+\mathrm{k}.3^{\mathrm{k}}+(\mathrm{k}+1)3^{\mathrm{k}+1}=\frac{\{2(\mathrm{k}+1)-1\}{3}^{\mathrm{k}+2}+3}{4}\\ \Rightarrow \frac{(2\mathrm{k}-1)3^{\mathrm{k}+1}+3}{4}+(\mathrm{k}+1)3^{\mathrm{k}+1}=\frac{(2\mathrm{k}+1)3^{\mathrm{k}+2}+3}{4}\text{\hspace{0.17em} …}\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{(2\mathrm{k}-1)3^{\mathrm{k}+1}+3}{4}+(\mathrm{k}+1)3^{\mathrm{k}+1}=\frac{(2\mathrm{k}-1)3^{\mathrm{k}+1}+3}{4}+(\mathrm{k}+1)3^{\mathrm{k}+1}\\ =\frac{(2\mathrm{k}-1)3^{\mathrm{k}+1}+3+4(\mathrm{k}+1)3^{\mathrm{k}+1}}{4}\\ =\frac{\{2\mathrm{k}-1+4(\mathrm{k}+1)\}{3}^{\mathrm{k}+1}+3}{4}\\ =\frac{\{2\mathrm{k}-1+4\mathrm{k}+4\}{3}^{\mathrm{k}+1}+3}{4}\\ =\frac{\{6\mathrm{k}+3\}{3}^{\mathrm{k}+1}+3}{4}\\ =\frac{(2\mathrm{k}+1)3^{\mathrm{k}+2}+3}{4}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.6 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{+}\mathbf{\dots }\mathbf{+}\mathbf{n}\mathbf{.}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}{\mathbf{3}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):\mathrm{}1.2+2.3+3.4+\dots +\mathrm{n}.(\mathrm{n}+1)=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1.2=2\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{1(1+1)(1+2)}{3}=\frac{1\left(2\right)\left(3\right)}{3}=2\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.2+2.3+3.4+\dots +\mathrm{k}.(\mathrm{k}+1)=\frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}{3}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.2+2.3+3.4+\dots +\mathrm{k}.(\mathrm{k}+1)+(\mathrm{k}+1)(\mathrm{k}+2)\\ =\frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}{3}\\ \Rightarrow \frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}{3}+(\mathrm{k}+1)(\mathrm{k}+2)=\frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}{3}...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}{3}+(\mathrm{k}+1)(\mathrm{k}+2)=(\mathrm{k}+1)(\mathrm{k}+2)(\frac{\mathrm{k}}{3}+1)\\ =\frac{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}{3}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.7 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{4}{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{3}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):\mathrm{}1.3+3.5+5.7+...+(2\mathrm{n}-1)(2\mathrm{n}+1)=\frac{\mathrm{n}(4{\mathrm{n}}^2+6\mathrm{n}-1)}{3}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=1.3=3\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{1.(4\times 1^2+6\times 1-1)}{3}=\frac{1\left(9\right)}{3}\\ =3\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.3+3.5+5.7+...+(2\mathrm{k}-1)(2\mathrm{k}+1)\\ =\frac{\mathrm{k}(4{\mathrm{k}}^2+6\mathrm{k}-1)}{3}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.3+3.5+5.7+...+(2\mathrm{k}-1)(2\mathrm{k}+1)+(2\mathrm{k}+1)(2\mathrm{k}+3)\\ =\frac{(\mathrm{k}+1)\{4{(\mathrm{k}+1)}^2+6(\mathrm{k}+1)-1\}}{3}\\ \Rightarrow \frac{\mathrm{k}(4{\mathrm{k}}^2+6\mathrm{k}-1)}{3}+(2\mathrm{k}+1)(2\mathrm{k}+3)=\frac{(\mathrm{k}+1)\{4({\mathrm{k}}^2+2\mathrm{k}+1)+6\mathrm{k}+6-1\}}{3}\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \Rightarrow \frac{\mathrm{k}(4{\mathrm{k}}^2+6\mathrm{k}-1)}{3}+(2\mathrm{k}+1)(2\mathrm{k}+3)=\frac{(\mathrm{k}+1)(4{\mathrm{k}}^2+8\mathrm{k}+4+6\mathrm{k}+6-1)}{3}\\ \Rightarrow \frac{\mathrm{k}(4{\mathrm{k}}^2+6\mathrm{k}-1)}{3}+(2\mathrm{k}+1)(2\mathrm{k}+3)=\frac{(\mathrm{k}+1)(4{\mathrm{k}}^2+14\mathrm{k}+9)}{3}...\left(2\right)\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\mathrm{k}(4{\mathrm{k}}^2+6\mathrm{k}-1)}{3}+(2\mathrm{k}+1)(2\mathrm{k}+3)=\frac{\mathrm{k}(4{\mathrm{k}}^2+6\mathrm{k}-1)}{3}+4{\mathrm{k}}^2+6\mathrm{k}+2\mathrm{k}+3\\ =\frac{4{\mathrm{k}}^3+6{\mathrm{k}}^2-\mathrm{k}+12{\mathrm{k}}^2+18\mathrm{k}+6\mathrm{k}+9}{3}\\ =\frac{4{\mathrm{k}}^3+18{\mathrm{k}}^2+23\mathrm{k}+9}{3}\\ =\frac{(\mathrm{k}+1)(4{\mathrm{k}}^2+14\mathrm{k}+9)}{3}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.8 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{+}\mathbf{2}\mathbf{.}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{.}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\mathbf{n}\mathbf{.}{\mathbf{2}}^{\mathbf{n}}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{–}\mathbf{1}\mathbf{)}{\mathbf{2}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):\mathrm{}1.2+2{.2}^2+3{.2}^2+...+\mathrm{n}.2^{\mathrm{n}}=(\mathrm{n}-1)2^{\mathrm{n}+1}+2\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right):\mathrm{L}.\mathrm{H}.\mathrm{S}.=1.2=2\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:(1-1)2^{1+1}+2=2\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):1.2+2{.2}^2+3{.2}^2+...+\mathrm{k}.2^{\mathrm{k}}=(\mathrm{k}-1)2^{\mathrm{k}+1}+2...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ 1.2+2{.2}^2+3{.2}^2+...+\mathrm{k}.2^{\mathrm{k}}+(\mathrm{k}+1).2^{\mathrm{k}+1}=(\mathrm{k}+1-1)2^{\mathrm{k}+1+1}+2\\ \Rightarrow (\mathrm{k}-1)2^{\mathrm{k}+1}+2+(\mathrm{k}+1).2^{\mathrm{k}+1}=\mathrm{k}.2^{\mathrm{k}+2}+2...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:(\mathrm{k}-1)2^{\mathrm{k}+1}+2+(\mathrm{k}+1).2^{\mathrm{k}+1}=(\mathrm{k}-1+\mathrm{k}+1)2^{\mathrm{k}+1}+2\\ =\left(2\mathrm{k}\right)2^{\mathrm{k}+1}+2\\ =\mathrm{k}.2^{\mathrm{k}+2}+2\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.9 Prove by using principal of mathematical induction for all n ∈ N

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{2}}^{\mathbf{n}}}\mathbf{=}\mathbf{1}\mathbf{–}\frac{\mathbf{1}}{{\mathbf{2}}^{\mathbf{n}}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):\mathrm{}\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{\mathrm{n}}}=1-\frac{1}{2^{\mathrm{n}}}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=\frac{1}{2}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:1-\frac{1}{2^1}=\frac{1}{2}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{\mathrm{k}}}=1-\frac{1}{2^{\mathrm{k}}}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^{\mathrm{k}}}+\frac{1}{2^{\mathrm{k}+1}}=1-\frac{1}{2^{\mathrm{k}+1}}\\ \Rightarrow 1-\frac{1}{2^{\mathrm{k}}}+\frac{1}{2^{\mathrm{k}+1}}=1-\frac{1}{2^{\mathrm{k}+1}}...\left(2\right)\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:1-\frac{1}{2^{\mathrm{k}}}+\frac{1}{2^{\mathrm{k}+1}}=1-\left(\frac{2-1}{2^{\mathrm{k}+1}}\right)\\ =1-\frac{1}{2^{\mathrm{k}+1}}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.10 Prove by using principal of mathematical induction for all n ∈ N

$\frac{\mathbf{1}}{\mathbf{2}\mathbf{.}\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{5}\mathbf{.}\mathbf{8}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{8}\mathbf{.}\mathbf{11}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{\mathbf{(}\mathbf{3}\mathbf{n}\mathbf{–}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{(}\mathbf{6}\mathbf{n}\mathbf{+}\mathbf{4}\mathbf{)}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\\ \mathrm{i}.\mathrm{e}.,\mathrm{P}\left(\mathrm{n}\right):\mathrm{}\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\mathrm{n}–1)(3\mathrm{n}+2)}=\frac{\mathrm{n}}{(6\mathrm{n}+4)}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right):\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1}{2.5}=\frac{1}{10}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{1}{(6\times 1+4)}=\frac{1}{10}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,}\\ \text{i.e., P}\left(\mathrm{k}\right):\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\mathrm{k}-1)(3\mathrm{k}+2)}=\frac{\mathrm{k}}{(6\mathrm{k}+4)}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3\mathrm{k}-1)(3\mathrm{k}+2)}+\frac{1}{\{3(\mathrm{k}+1)-1\}\{3(\mathrm{k}+1)+2\}}\\ =\frac{\mathrm{k}+1}{\{6(\mathrm{k}+1)+4\}}\\ \Rightarrow \frac{\mathrm{k}}{(6\mathrm{k}+4)}+\frac{1}{(3\mathrm{k}+2)(3\mathrm{k}+5)}=\frac{\mathrm{k}+1}{(6\mathrm{k}+10)}...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{\mathrm{k}}{(6\mathrm{k}+4)}+\frac{1}{(3\mathrm{k}+2)(3\mathrm{k}+5)}=\frac{3{\mathrm{k}}^2+5\mathrm{k}+2}{2(3\mathrm{k}+2)(3\mathrm{k}+5)}\\ =\frac{(3\mathrm{k}+2)(\mathrm{k}+1)}{2(3\mathrm{k}+2)(3\mathrm{k}+5)}\\ =\frac{\mathrm{k}+1}{(6\mathrm{k}+10)}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.11 Prove by using principal of mathematical induction for all n ∈ N

$\frac{\mathbf{1}}{\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{.}\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{.}\mathbf{4}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{.}\mathbf{5}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{)}}{\mathbf{4}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\mathrm{i}.\mathrm{e}.,\\ \mathrm{P}\left(\mathrm{n}\right):\mathrm{}\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}=\frac{\mathrm{n}(\mathrm{n}+3)}{4(\mathrm{n}+1)(\mathrm{n}+2)}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right):\frac{1}{1.2.3}=\frac{1}{6}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{1(1+3)}{4(1+1)(1+2)}=\frac{4}{4\times 2\times 3}=\frac{1}{6}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k, i.e.,}\\ \text{We have}\\ \text{P}\left(\mathrm{k}\right):\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}\\ =\frac{\mathrm{k}(\mathrm{k}+3)}{4(\mathrm{k}+1)(\mathrm{k}+2)}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\mathrm{k}(\mathrm{k}+1)(\mathrm{k}+2)}+\frac{1}{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}\\ =\frac{(\mathrm{k}+1)(\mathrm{k}+4)}{4(\mathrm{k}+2)(\mathrm{k}+3)}\\ \Rightarrow \frac{\mathrm{k}(\mathrm{k}+3)}{4(\mathrm{k}+1)(\mathrm{k}+2)}+\frac{1}{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}\\ =\frac{(\mathrm{k}+1)(\mathrm{k}+4)}{4(\mathrm{k}+2)(\mathrm{k}+3)}...\left(2\right)\left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{\mathrm{k}(\mathrm{k}+3)}{4(\mathrm{k}+1)(\mathrm{k}+2)}+\frac{1}{(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}\\ =\frac{\mathrm{k}{(\mathrm{k}+3)}^2+4}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}\\ =\frac{\mathrm{k}({\mathrm{k}}^2+6\mathrm{k}+9)+4}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}\\ =\frac{{\mathrm{k}}^3+6{\mathrm{k}}^2+9\mathrm{k}+4}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}\\ =\frac{(\mathrm{k}+1)({\mathrm{k}}^2+5\mathrm{k}+4)}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}\\ =\frac{(\mathrm{k}+1)(\mathrm{k}+1)(\mathrm{k}+4)}{4(\mathrm{k}+1)(\mathrm{k}+2)(\mathrm{k}+3)}\\ =\frac{(\mathrm{k}+1)(\mathrm{k}+4)}{4(\mathrm{k}+2)(\mathrm{k}+3)}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.12 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{a}\mathbf{+}\mathbf{ar}\mathbf{+}{\mathbf{ar}}^{\mathbf{2}}\mathbf{+}{\mathbf{ar}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathbf{ar}}^{\mathbf{n}\mathbf{–}\mathbf{1}}\mathbf{=}\frac{\mathbf{a}\mathbf{(}{\mathbf{r}}^{\mathbf{n}}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{r}\mathbf{-}\mathbf{1}}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\mathrm{i}.\mathrm{e}.,\\ \mathrm{P}\left(\mathrm{n}\right):\mathrm{}\mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^2+{\mathrm{ar}}^3+...+{\mathrm{ar}}^{\mathrm{n}–1}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{n}}-1)}{\mathrm{r}-1}\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right):\mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{a}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:\frac{\mathrm{a}({\mathrm{r}}^1-1)}{\mathrm{r}-1}=\mathrm{a}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,i.e.,}\\ \text{P}\left(\mathrm{k}\right):\mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^2+{\mathrm{ar}}^3+...+{\mathrm{ar}}^{\mathrm{k}–1}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{k}}-1)}{\mathrm{r}-1}...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ \mathrm{a}+\mathrm{ar}+{\mathrm{ar}}^2+{\mathrm{ar}}^3+...+{\mathrm{ar}}^{\mathrm{k}–1}+{\mathrm{ar}}^{\mathrm{k}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{k}+1}-1)}{\mathrm{r}-1}\\ \Rightarrow \frac{\mathrm{a}({\mathrm{r}}^{\mathrm{k}}-1)}{\mathrm{r}-1}+{\mathrm{ar}}^{\mathrm{k}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{k}+1}-1)}{\mathrm{r}-1}...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{k}}-1)}{\mathrm{r}-1}+{\mathrm{ar}}^{\mathrm{k}}=\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{k}}-1)+{\mathrm{ar}}^{\mathrm{k}}(\mathrm{r}-1)}{\mathrm{r}-1}\\ =\frac{{\mathrm{ar}}^{\mathrm{k}}-\mathrm{a}+{\mathrm{ar}}^{\mathrm{k}+1}-{\mathrm{ar}}^{\mathrm{k}}}{\mathrm{r}-1}\\ =\frac{\mathrm{a}({\mathrm{r}}^{\mathrm{k}+1}-1)}{\mathrm{r}-1}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.13 Prove by using principal of mathematical induction for all n ∈ N

$(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})...(1+\frac{(2\mathrm{n}+1)}{{\mathrm{n}}^2})={(\mathrm{n}+1)}^2.$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\mathrm{i}.\mathrm{e}.,\\ \mathrm{P}\left(\mathrm{n}\right):(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})...\{1+\frac{(2\mathrm{n}+1)}{{\mathrm{n}}^2}\}={(\mathrm{n}+1)}^2\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=(1+\frac{3}{1})=4\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:{(1+1)}^2=4\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,i.e.,}\\ \text{P}\left(\mathrm{k}\right):(1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})...\{1+\frac{(2\mathrm{k}+1)}{{\mathrm{k}}^2}\}={(\mathrm{k}+1)}^2\text{\hspace{0.17em} …}\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ (1+\frac{3}{1})(1+\frac{5}{4})(1+\frac{7}{9})...\{1+\frac{(2\mathrm{k}+1)}{{\mathrm{k}}^2}\}\{1+\frac{(2\mathrm{k}+3)}{{(\mathrm{k}+1)}^2}\}={(\mathrm{k}+2)}^2\\ \Rightarrow {(\mathrm{k}+1)}^2\{1+\frac{(2\mathrm{k}+3)}{{(\mathrm{k}+1)}^2}\}={(\mathrm{k}+2)}^2\text{\hspace{0.17em} …}\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(1\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:{(\mathrm{k}+1)}^2\{1+\frac{(2\mathrm{k}+3)}{{(\mathrm{k}+1)}^2}\}={(\mathrm{k}+1)}^2\left\{\frac{{(\mathrm{k}+1)}^2+(2\mathrm{k}+3)}{{(\mathrm{k}+1)}^2}\right\}\\ ={\mathrm{k}}^2+2\mathrm{k}+1+2\mathrm{k}+3\\ ={\mathrm{k}}^2+4\mathrm{k}+4\\ ={(\mathrm{k}+2)}^2\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$

Q.14 Prove by using principal of mathematical induction for all n ∈ N

$\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{)}\mathbf{.}\mathbf{..}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{Let}\text{P}\left(\mathrm{n}\right)\text{be the given statement,}\mathrm{i}.\mathrm{e}.,\\ \mathrm{P}\left(\mathrm{n}\right):(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{\mathrm{n}})=(\mathrm{n}+1)\\ \mathrm{For}\text{n}=1,\\ \mathrm{P}\left(1\right)=(1+\frac{1}{1})=2\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.:(1+1)=2\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{P}\left(\mathrm{n}\right)\text{\hspace{0.17em}is true for n}=1.\\ \text{Let P(k) is true for some natural number k,i.e.,}\\ \text{P}\left(\mathrm{k}\right):(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{\mathrm{k}})=(\mathrm{k}+1)...\left(1\right)\\ \text{We need to prove that}\mathrm{P}(\mathrm{k}+1)\text{is true whenever P(k) is true.}\\ \text{We have}\\ (1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{\mathrm{k}})(1+\frac{1}{\mathrm{k}+1})=(\mathrm{k}+2)\\ \Rightarrow (\mathrm{k}+1)(1+\frac{1}{\mathrm{k}+1})=(\mathrm{k}+2)...\left(2\right)\\ \left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:(\mathrm{k}+1)(1+\frac{1}{\mathrm{k}+1})=(\mathrm{k}+1)\left(\frac{\mathrm{k}+2}{\mathrm{k}+1}\right)\\ =(\mathrm{k}+2)\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \text{Therefore, P(k + 1) is true when P(k) is true. Hence, by}\\ \text{principle of mathematical induction, P(n) is true for every}\\ \text{positive integer n.}\end{array}$