Question

Prove : $1^{+}{2}^2+3^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ by principal of Mathematical Induction.

Answer 1

For, $n=1$ the statement reduces to

$1^2=\frac{\mathrm{1.2.3}}{6}$ and it is true

Let us assume that the statement is true for $n=k$

i.e, $1^2+2^2+3^2+.....+k^2=\frac{k(k+1)(2k+1)}{6}⟶\left(1\right)$

Now we have to prove that the statement is true for $n=k+1$

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

BY principle of mathematical induction the given statement is true for every positive integer n.