Prove that 1+4+9+...+n^2 = n(n+1)(2n+1)/6.

Consider the case n=1. Then 1(1+1)(2*1+1)/6 = 1 = 1^2 and so the claim is true for n=1. Suppose the claim is true for some positive integer n, so that 1+4+9+...+n^2 = n(n+1)(2n+1)/6. Then by the inductive hypothesis 1+4+9+...+n^2 + (n+1)^2 = (1+4+9+...+n^2) + (n+1)^2 = n(n+1)(2n+1)/6 + (n+1)^2                                                                                      = (n+1)(2n^2 + n + 6(n+1))/6                                                                                      = (n+1)(n+2)(2n+3)/6 which is the claim for n+1. As the claim is true for n=1, it's true for all n by induction.