# Finding a Suitable Telescoping Sum to Use in Problem [duplicate]

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!

• hint: $1^3=1^2-0^2$, $2^3=3^2-1^2$, $3^3=6^2-3^2$, ... Sep 27 '18 at 18:25
• See the telescoping sum used by 1233dfv in his answer at the duplicate. It comes from $\sum_k ((k+1)^4-k^4)$. Sep 27 '18 at 19:44

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)

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

• Where did that identity come from? Sep 27 '18 at 19:40
• From the complete square formula. Sep 27 '18 at 19:51