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