For all $n \ge 1,$ prove that $1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$

(N/A) Let the given statement be $P(n),$ i.e.,
$P(n): 1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2n+1)}{6}$
For $n=1, P(1): 1^{2} = \frac{1(1+1)(2 \times 1+1)}{6} = \frac{1 \times 2 \times 3}{6} = 1,$ which is true.
Assume that $P(k)$ is true for some positive integer $k,$ i.e.,
$1^{2}+2^{2}+\ldots+k^{2}=\frac{k(k+1)(2k+1)}{6} \quad \dots(1)$
We shall now prove that $P(k+1)$ is also true.
Consider the sum of the first $(k+1)$ terms:
$(1^{2}+2^{2}+\ldots+k^{2})+(k+1)^{2} = \frac{k(k+1)(2k+1)}{6} + (k+1)^{2} \quad [\text{Using } (1)]$
$= \frac{(k+1)[k(2k+1) + 6(k+1)]}{6}$
$= \frac{(k+1)[2k^{2}+k+6k+6]}{6}$
$= \frac{(k+1)[2k^{2}+7k+6]}{6}$
$= \frac{(k+1)(k+2)(2k+3)}{6}$
$= \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \ge 1.$