Well, we can prove that
∑0→n i3 = (n4 + 2n3 + n2) / 4
(Note that the last term is different from that of the question, which has a typo!)
But first we express the right-hand side of the equation differently by factoring it (because factored forms are likely to be easier to manipulate algebraically). Thus,
∑0→n i3 = (n4 + 2n3 + n2) / 4
= n2(n2 + 2n + 1) / 4
= n2(n + 1)2 / 4
Thus, we instead prove the following.
∑0→n i3 = n2(n + 1)2 / 4
Proof by Mathematical Induction
Base case. For the case of n = 0, the left-hand side of the equation is ∑0→n i3 = 03 = 0, and the right-hand side is n2(n + 1)2 / 4 = 02(0 + 1)2 / 4 = 0 / 4 = 0, which makes the equation a true statement.
Inductive step. For the case of n = k + 1, the left-hand side is
∑0→k+1 i3 = (k+1)3 + ∑0→k i3
in which we can substitute the inductive hypothesis ∑0→k i3 = k2(k + 1)2 / 4. Thus,
∑0→k+1 i3 = (k+1)3 + k2(k + 1)2 / 4
= (4(k+1)3 + k2(k + 1)2) / 4
= (k+1)2(4(k + 1) + k2) / 4
= (k+1)2(k2 + 4k + 4) / 4
= (k+1)2(k + 2)2 / 4
= (k+1)2((k+1) + 1)2 / 4
which completes the inductive step and the induction proof. 〈