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I need to prove that

$$\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$

The base case is true, so after assuming it's true for n, I proceeded to the inductive step:

$$\sum_{i=1}^{n+1} i^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$$

That's where I wanna get. Then:

$$\sum_{i=1}^{n+1} i^3 = \sum_{i=1}^n i^3 + (n+1)^3 = \left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3$$

And this is where I'm stuck at. I couldn't find a way to merge both terms into one as they have different exponents. So I can't do something like $\frac{((n(n+1))^2+2(n+1)^3}{2}$, right?

I also tried doing $\sum_{i=1}^{n+1} i^3 = \sum_{i=1}^n i^2 . \sum_{i=1}^n i$ but couldn't try to make both expressions into what I'm looking for.

Thanks in advance.

Ethan Bolker
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David Zavalla
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    Check this out https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct – K Split X Apr 23 '17 at 23:16

1 Answers1

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They do have different exponents, but you can definitely merge them by factorizing.

$$\sum_{i=1}^{n+1} i^3 = \sum_{i=1}^n i^3 + (n+1)^3 = \left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3=(n+1)^2\left( \frac{n^2}{4}+(n+1)\right)=(n+1)^2\left( \frac{n^2+4n+4}{4}\right)=(n+1)^2\left( \frac{(n+2)^2}{4}\right)$$$$=\left(\frac{(n+1)(n+2)}{2}\right)^2$$

as desired.

Jaideep Khare
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