# 12 + 22 + 32 + ... + n

Question:

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

Solution:

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 $n \in N$.