Prove by induction: $\sum_{i=1}^n i^3 = \left[\sum_{i=1}^n i\right]^2$. Hint: Use $k(k+1)^2 = 2(k+1)\sum i$.

Basis: $n = 1$ $\sum_{i=1}^1 i^3 = \left[\sum_{i=1}^1 i\right]^2 \to 1^3 = 1^2 \to 1 = 1$.

Hypothesis: Assume true for all $n \le k$.

So far I have the following:

$$\sum_{i=1}^{k+1} i^3 = (k+1)^3 + \sum_{i=1}^k i^3$$

$$(k+1)^3 + \left[\sum_{i=1}^k i\right]^2$$

  • $\begingroup$ you're nearly there. try fiddling with the $(k+1)^3$ piece on the left a bit more. Also, while a final and rigorous proof won't do it, you might try working backwards instead, since the square of the sum is harder to work with than the sum of the cubes. To finish this, it will require you to expand the square of the sum, add the extra bit from the piece on the left, and then put it back together again I expect. $\endgroup$ – JMoravitz Oct 29 '14 at 3:00
  • $\begingroup$ See math.stackexchange.com/questions/1111443/…, math.stackexchange.com/questions/62171/… and other posts you can find linked there. $\endgroup$ – Martin Sleziak Feb 16 '15 at 13:16

For $n=k+1$, $$\sum_{i=1}^{k+1}i^3 = \sum_{i=1}^{k}i^3+(k+1)^3=(\sum_{i=1}^{k}i)^2+(k+1)^3=(\sum_{i=1}^{k}i)^2+k(k+1)^2+(k+1)^2$$

Now using the Hint: $k(k+1)^2 = 2(k+1)\sum i$.

$$=(\sum_{i=1}^{k}i)^2+2(k+1)\sum_{i=1}^k i+(k+1)^2=(\sum_{i=1}^{k+1}i)^2$$

  • $\begingroup$ Hi Paul, Thank you for the help! Could you expand on how you turned $(k +1)^3$ into $k(k+1)^2+(k+1)^2$? I'm multiplying everything right now hoping to get the same result, but I'm ending up with $k^3 + 3k^2 + 3k + 1$ $\endgroup$ – mylasthope Oct 29 '14 at 3:06
  • $\begingroup$ Hi @mylasthope: $(k+1)^3=(k+1)\times (k+1)^2=k\times (k+1)^2+1\times (k+1)^2.$ $\endgroup$ – Paul Oct 29 '14 at 3:15

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