Principle of Mathy Induction.........

**kodirl** · August 9th 2006, 06:31 AM

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)

**galactus** · August 9th 2006, 07:21 AM

Presume $P_{1}$ is true, since $\frac{(1)(1+1)(2(1)+1)}{6}=1$

Presume $P_{k}$ is true:

$1^{2}+2^{2}+3^{2}+...........+k^{2}=\frac{k(k+1)(2 k+1)}{6}$

Therefore,

$1^{2}+2^{2}+3^{2}+............+k^{2}+(k+1)^{2}$ = $\frac{k(k+1)(2k+1)}{6}+(k+1)^{2}$

= $(k+1)\left[\frac{k(2k+1)}{6}+\frac{6(k+1)}{6}\right]$

= $\frac{(k+1)(2k^{2}+7k+6}{6})$

= $\frac{(k+1)(k+2)(2k+3)}{6}$

So, $P_{k+1}$ is true. QED

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