State the principle of mathematical induction and use it to prove that;
1^2+2^2+3^2+...+n^2=1/6n(n+1)(2n+1)
Presume $\displaystyle P_{1}$ is true, since $\displaystyle \frac{(1)(1+1)(2(1)+1)}{6}=1$
Presume $\displaystyle P_{k}$ is true:
$\displaystyle 1^{2}+2^{2}+3^{2}+...........+k^{2}=\frac{k(k+1)(2 k+1)}{6}$
Therefore,
$\displaystyle 1^{2}+2^{2}+3^{2}+............+k^{2}+(k+1)^{2}$=$\displaystyle \frac{k(k+1)(2k+1)}{6}+(k+1)^{2}$
=$\displaystyle (k+1)\left[\frac{k(2k+1)}{6}+\frac{6(k+1)}{6}\right]$
=$\displaystyle \frac{(k+1)(2k^{2}+7k+6}{6})$
=$\displaystyle \frac{(k+1)(k+2)(2k+3)}{6}$
So, $\displaystyle P_{k+1}$ is true. QED