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Prove that $1+4+9+16+...+n^2= \frac{1}{6}n(n+1)(2n+1)$,

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It can be proved by mathematical induction. Let p(k) be 1+4.. +k²=1/6{k(k+1)( 2k+1)} P(k+1) = 1+4+...+k²+(k+1) ² =1/6{k(k+1) (2k+1) } +(k+1) ² Solving the above equation, we get P(k+1) =1/6{(k+1) (k+2) (2k+3) } Thus, the above equation is true for all n (natural numbers).

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