Browse Questions

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

Let the given statement be $P(n)$, i.e.,
$P(n) :1^2+2^2+3^2+4^4+...+n^2=\large\frac{n(n+1)(2n+1)}{6}$.
For $n=1$, $P(1): 1 = \large\frac{1(1+1)(2 \times 1+1)}{6}$$= \large\frac{1 \times 2 \times 3 }{6}$$=1$ which is true.
Assume that $P(k)$ is true for some positive integers $k$, i.e.,
$1^2+2^2+3^2+4^4+...+k^2=\large\frac{k(k+1)(2k+1)}{6}$----------(1)
We shall now prove that $P(k+1)$ is also true. Now, we have
$(1^2+2^2+3^2+4^2...+n^2)+(k+1)^2$