Question

1² + 2² + 3² + ... + n² = $\frac{n(n+1)(2n+1)}{6}$

Answer 1

Let P(n) be the given statement.
Now,

$$\begin{gathered} P\left(n\right)=1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6} \\ \mathrm{Step}1: \\ P\left(1\right)=1^2=\frac{1(1+1)(2+1)}{6}=\frac{6}{6}=1 \\ \mathrm{Hence},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)\mathrm{be}\mathrm{true}. \\ \mathrm{Then}, \\ {1}^2+2^2+...+m^2=\frac{m(m+1)(2m+1)}{6} \\ \mathrm{We}\mathrm{shall}\mathrm{now}\mathrm{prove}\mathrm{that}P(m+1)\mathrm{is}\mathrm{true}. \\ \mathrm{i}.\mathrm{e}., \\ {1}^2+2^2+3^2+...+(m+1{)}^2=\frac{(m+1)(m+2)(2m+3)}{6} \\ \mathrm{Now}, \\ P(m)=1^2+2^2+3^2+...+m^2=\frac{m(m+1)(2m+1)}{6} \\ \Rightarrow 1^2+2^2+3^2+...+m^2+(m+1{)}^2=\frac{m(m+1)(2m+1)}{6}+(m+1{)}^2\left[\mathrm{Adding}(m+1{)}^2\mathrm{to}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow 1^2+2^2+3^2+...+(m+1{)}^2=\frac{m(m+1)(2m+1)+6(m+1{)}^2}{6}=\frac{(m+1)(2m^2+m+6m+6)}{6}=\frac{(m+1)(m+2)(2m+3)}{6} \\ \mathrm{Hence},P(m+1)\mathrm{is}\mathrm{true}. \\ Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}in\mathrm{duction},\mathrm{the}\mathrm{given}\mathrm{statement}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N. \end{gathered}$$