sum($k^2$, k=1 to n) = n(n+1)$\frac{2n+1}{6}$. This can be proved by induction or by using the identity $k^{3-}$(k-1)^3 = 3$k^{2-3k+1}$ (telescoping). For n=10: 101121/6 = 385. Remember the factor pattern: n(n+1) multiplied by $\frac{2n+1}{6}$. For sum of squares of first n even numbers: sum((2k)^2) = 4*sum($k^2$) = 2n(n+1)$\frac{2n+1}{3}$.