Question

$1^2+2^2+3^2+......+n^2=\frac{n(n+1)(2n+1)}{6}$

Answer 1

Let P(n) : $1^2+2^2+3^2+.......+n^2=\frac{n(n+1)(2n+1)}{6}$

For $n=1$

p(1) : 1 = 1 $\frac{1(1+1)(2+1)}{6}$

$1=1$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

$p\left(k\right):1^2+2^2+3^2+......+k^2=\frac{k(k+1)(2k+1)}{6}$ .......(1)

We have to show that p(n) is true for n = k + 1

$\Rightarrow 1^2+2^2+3^2+.....+k^2+(k+1{)}^2=\frac{\left(\right)(k+1)(k+2)(2k+3)}{6}$

So, $1^2+2^2+3^2+......+k^2+(k+1{)}^2$.

$=\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2$

[Using equation (1)]

$=(k+1)[\frac{2k^2+k}{6}+\frac{(k+1)}{1}]$

$=(k+1)\left[\frac{2k^2+k+6k+6}{6}\right]$

$=(k+1)\left[\frac{2k^2+7k+6}{6}\right]$

$=(k+1)\left[\frac{2k^2+4k+3k+6}{6}\right]$

$=(k+1)\left[\frac{2k(k+2)+3(k+2)}{6}\right]$

$=\frac{(k+1)(2k+3)(k+2)}{6}$

$\Rightarrow$ P(n) is true for n = k + 1

$\Rightarrow$ P(n) is true for all $nϵN$ by PMI