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I am currently lost on the following problem:

Use a telescoping sum to give a proof without induction that for each $n \in \mathbb{N},$

$$1^3+2^3+3^3+\dots +n^3=\frac{n^2(n+1)^2}{4}$$

I have followed other examples where they show equalities like this, but I don't understand how they seem to come up with the telescoping series they use to solve them.

Thank you!

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    $\begingroup$ hint: $1^3=1^2-0^2$, $2^3=3^2-1^2$, $3^3=6^2-3^2$, ... $\endgroup$
    – Vasya
    Sep 27 '18 at 18:25
  • $\begingroup$ See the telescoping sum used by 1233dfv in his answer at the duplicate. It comes from $\sum_k ((k+1)^4-k^4)$. $\endgroup$ Sep 27 '18 at 19:44
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For all $k \in \mathbb{N}$, you have $$(k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$$

So $$(k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1$$

Summing for $k=1$ to $n$, you get $$(n+1)^4-1=\sum_{k=1}^n (4k^3 + 6k^2 + 4k + 1)$$

So using the well-known values of $\sum_{k=1}^n k^2$ and $\sum_{k=1}^n k$, $$\sum_{k=1}^n k^3 = \frac{1}{4}((n+1)^4-1 - n(n+1)(2n+1) - 2n(n+1) - n)$$

i.e. $$\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$$

(If the values of $\sum_{k=1}^n k^2$ and $\sum_{k=1}^n k$ are not "well-known", you can actually compute them with the same process)

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Using the identity $$ k^3=\frac14\bigg[k^2(k+1)^2-(k-1)^2k^2\bigg] $$ it is easy to see $$ \sum_{k=1}^n k^3=\frac14\sum_{k=1}^n\bigg[k^2(k+1)^2-(k-1)^2k^2\bigg]=\frac14n^2(n+1)^2. $$

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  • $\begingroup$ Where did that identity come from? $\endgroup$
    – M. Damon
    Sep 27 '18 at 19:40
  • $\begingroup$ From the complete square formula. $\endgroup$
    – xpaul
    Sep 27 '18 at 19:51

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