Question

Show by the Principle of Mathematical induction that the sum S_n of then terms of the series $1^2+2\times 2^2+3^2+2\times 4^2+5^2+2\times 6^2+7^2+...$ is given by

$S_n=\left\{\begin{array}{l}\frac{n{\left(n+1\right)}^2}{2},\mathrm{if}n\mathrm{is}\mathrm{even}\\ \frac{n^2\left(n+1\right)}{2},\mathrm{if}n\mathrm{is}\mathrm{odd}\end{array}\right.$ [NCERT EXEMPLAR]

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{P}\left(n\right):S_n=1^2+2\times 2^2+3^2+2\times 4^2+5^2+...=\left\{\begin{array}{l}\frac{n{\left(n+1\right)}^2}{2},\mathrm{when}n\mathrm{is}\mathrm{even}\\ \frac{n^2\left(n+1\right)}{2},\mathrm{when}n\mathrm{is}\mathrm{odd}\end{array}\right. \\ \mathrm{Step}\mathrm{I}:\mathrm{For}n=1\mathrm{i}.\mathrm{e}.\mathrm{P}\left(1\right): \\ \mathrm{LHS}={\mathrm{S}}_1=1^2=1 \\ \mathrm{RHS}={\mathrm{S}}_1=\frac{1^2\left(1+1\right)}{2}=1 \\ \mathrm{As},\mathrm{LHS}=\mathrm{RHS} \\ \mathrm{So},\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}n=1. \\ \mathrm{Step}\mathrm{II}:\mathrm{For}n=k, \\ \mathrm{Let}\mathrm{P}\left(k\right):S_k=1^2+2\times 2^2+3^2+2\times 4^2+5^2+...=\left\{\begin{array}{l}\frac{k{\left(k+1\right)}^2}{2},\mathrm{when}k\mathrm{is}\mathrm{even}\\ \frac{k^2\left(k+1\right)}{2},\mathrm{when}k\mathrm{is}\mathrm{odd}\end{array}\right.,\mathrm{be}\mathrm{true}\mathrm{for}\mathrm{some}\mathrm{natural}\mathrm{numbers}. \\ \mathrm{Step}\mathrm{III}:\mathrm{For}n=k+1, \\ \mathrm{Case}1:\mathrm{When}k\mathrm{is}\mathrm{odd},\mathrm{then}\left(k+1\right)\mathrm{is}\mathrm{even}. \\ \mathrm{P}\left(k+1\right): \\ \mathrm{LHS}=S_{k+1}=1^2+2\times 2^2+3^2+2\times 4^2+5^2+...+k^2+2\times {\left(k+1\right)}^2 \\ =\frac{k^2\left(k+1\right)}{2}+2\times {\left(k+1\right)}^2\left(\mathrm{Using}\mathrm{step}\mathrm{II}\right) \\ =\frac{k^2\left(k+1\right)+4{\left(k+1\right)}^2}{2} \\ =\frac{\left(k+1\right)\left(k^2+4k+4\right)}{2} \\ =\frac{\left(k+1\right){\left(k+2\right)}^2}{2} \\ \mathrm{RHS}=\frac{\left(k+1\right){\left(k+1+1\right)}^2}{2} \\ =\frac{\left(k+1\right){\left(k+2\right)}^2}{2} \\ \mathrm{As},\mathrm{LHS}=\mathrm{RHS} \\ \mathrm{So},\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}n=k+1\mathrm{when}k\mathrm{is}\mathrm{odd}. \\ \mathrm{Case}2:\mathrm{When}k\mathrm{is}\mathrm{even},\mathrm{then}\left(k+1\right)\mathrm{is}\mathrm{odd}. \\ \mathrm{P}\left(k+1\right): \\ \mathrm{RHS}=S_{k+1}=1^2+2\times 2^2+3^2+2\times 4^2+5^2+...+2\times k^2+{\left(k+1\right)}^2 \\ =\frac{k{\left(k+1\right)}^2}{2}+{\left(k+1\right)}^2\left(\mathrm{Using}\mathrm{step}\mathrm{II}\right) \\ =\frac{k{\left(k+1\right)}^2+2{\left(k+1\right)}^2}{2} \\ =\frac{{\left(k+1\right)}^2\left(k+2\right)}{2} \\ =\frac{{\left(k+1\right)}^2\left(k+2\right)}{2} \\ \mathrm{RHS}=\frac{{\left(k+1\right)}^2\left(k+1+1\right)}{2} \\ =\frac{{\left(k+1\right)}^2\left(k+2\right)}{2} \\ \mathrm{As},\mathrm{LHS}=\mathrm{RHS} \\ \mathrm{So},\mathrm{it}\mathrm{is}\mathrm{true}\mathrm{for}n=k+1\mathrm{when}k\mathrm{is}\mathrm{even}. \\ \mathrm{Hence},\mathrm{P}\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}\mathrm{natural}\mathrm{numbers}. \end{gathered}$$