Question

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

Answer 1

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

for $n=1$

$LHS=1^2=1$

$RHS=\frac{\left(1\right)(1+1)(2\times 1+1)}{6}=\frac{1\times 2\times 3}{6}=1$

LHS = RHS

$P\left(n\right)$ is true for $n=1$

Assume that $P\left(k\right)$ is true

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

we will prove $P(k+1)$ is true

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

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

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

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

$=\frac{(k+1)(2k^2+7k+1)}{6}$

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

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

$\therefore P(k+1)$ is true when $p\left(k\right)$ is true

$\therefore$ BY principle of mathematical induction , $P\left(n\right)$ true for n where n is a natural number