Prove that for all $n \in \mathbb{Z}^+$, $\sum_{i=1}^n i^3 = \left(\sum_{i=1}^n i\right)^2$. I understand how to do the base case using the proof of induction, but I don't know how you would show the inductive step given that there is a summation on either side of the equal sign.
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Assume that $$\sum_{i=1}^k i^3 = (\sum_{i=1}^k i)^2.$$
We prove that $$\sum_{i=1}^{k+1} i^3 = (\sum_{i=1}^{k+1} i)^2.$$
Indeed, we have $$LHS = \sum_{i=1}^{k} i^3 + (k+1)^3 = (\sum_{i=1}^k i)^2 + (k+1)^3 = \frac{k^2(k+1)^2}{4} + (k+1)^3 = (k+1)^2\left[\frac{k^2}{4} + k+1\right] = (k+1)^2\frac{(k+2)^2}{4} = (\sum_{i=1}^{k+1} i)^2.$$
GAVD
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How did you get $(k^2 (k+1)^2)/4$ ? – Katie Loss Sep 14 '15 at 15:05
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Do you know $\sum_{i=1}^k i = \frac{k(k+1)}{2}$? – GAVD Sep 15 '15 at 02:02