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