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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}$.

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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
$(1^2+2^2+3^2+4^2...+n^2)+(k+1)^2$
$ \large\frac{k(k+1)(2k+1)}{6}$$+(k+1)^2\qquad$ [ Using (1) ]
$=\large\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$
$ = \large\frac{(k+1)(2k^2+7k+6)}{6}$
$ = \large\frac{(k+1)(k+1+1)\{2(k+1)+1\}}{6}$
Thus $P(k+1)$ is true, whenever $P(k)$ is true.
Hence, from the principle of mathematical induction, the statement $P(n)$ is true for all natural numbers $n$.
answered May 2, 2014 by thanvigandhi_1
 

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