Use mathematical induction to prove that 1^2 + 2^2 + 3^2.... + n^2 = (n(n+1)(2n+1))/6
thanks!
It's obviously true for n=1.
Assume $\displaystyle 1^{2}+2^{2}+3^{2}+k^{2} = \frac {k(k+1)(2k+1)}{6} $ is true for some k>1.
then $\displaystyle 1^{2}+2^{2}+3^{2}+k^{2}+(k+1)^{2} = \frac {k(k+1)(2k+1)}{6} + (k+1)^{2} $
$\displaystyle = \frac{2k^{3}+9k^{2}+13k+6}{6} $
$\displaystyle = \frac{(k+1)(2k^{2}+7k+6)}{6} $
$\displaystyle = \frac {(k+1)(k+2)(2k+3)}{6} $
$\displaystyle = \frac{\big(k+1\big)\big((k+1)+1\big)\big(2(k+1)+1\ big)}{6} $