(c) Prove that for any positive integer n, sum_{j=1}^{n} j(j - 1) = (n(n^2 - 1))/3
Solution
Proof.
By induction on n.
Base case: n = 1. sum_{j=1}^{1} j(j - 1) = 1 * (1 - 1) = 0 and (n(n^2 - 1))/3 = (1(1^2 - 1))/3 = 0, so equality holds when n = 1.
Inductive step: Assume that k is a positive integer and sum_{j=1}^{k} j(j - 1) = (k(k^2 - 1))/3.
Show that sum_{j=1}^{k+1} j(j - 1) = ((k+1)((k+1)^2 - 1))/3.
sum_{j=1}^{k+1} j(j - 1) = sum_{j=1}^{k} j(j - 1) + (k + 1)k
= (k(k^2 - 1))/3 + (k + 1)k by the inductive hypothesis
= (k(k^2 - 1) + 3(k + 1)k)/3
= (k(k + 1)(k - 1) + 3(k + 1)k)/3
= ((k + 1)[k(k - 1) + 3k])/3
= ((k + 1)[k^2 - k + 3k])/3
= ((k + 1)[k^2 + 2k])/3
= ((k + 1)[(k + 1)^2 - 1])/3
Therefore, sum_{j=1}^{k+1} j(j - 1) = ((k+1)((k+1)^2 - 1))/3.