Question 504679
It is easy to prove via induction; but more difficult to derive the formula. To prove it by induction, note that the base case n = 1 holds. Assume it holds for n=k, e.g.


*[tex \LARGE 1^2 + ... + k^2 = \frac{k(k+1)(2k+1)}{6}]. Then we want to show it works for k+1:


*[tex \LARGE 1^2 + ... + (k+1)^2 = (1 + ... + k^2) + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2]


*[tex \LARGE = \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6} = \frac{(k+1)(k(2k+1) + 6(k+1))}{6}]


*[tex \LARGE = \frac{(k+1)(2k^2 + 7k + 6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}]


which is what we would've gotten if we substituted k+1 into the formula. Hence the statement is true for all positive integers n.