Proving the sum of cubes?

Question

(a) Prove that for every natural number $n$, $$\sum_{k=1}^n k^3 = \frac14 n^2(n+1)^2.$$

(b) Can you think of any natural reason for that fact that this is equal to $\left(\sum_{k=1}^n k^2\right)$?

I'm really stuck as to where to start, I know I could probably google the first part but how does it relate to the second part?

Thanks in advance!

Answer 1

You can prove this by induction.

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First, show that this is true for $n=1$:

$\sum\limits_{k=1}^{1}k^3=1^2(1+1)^2/4$

Second, assume that this is true for $n$:

$\sum\limits_{k=1}^{n}k^3=n^2(n+1)^2/4$

Third, prove that this is true for $n+1$:

$\sum\limits_{k=1}^{n+1}k^3=$

$\color\red{\sum\limits_{k=1}^{n}k^3}+(n+1)^3=$

$\color\red{n^2(n+1)^2/4}+(n+1)^3=$

$(n+1)^2(n+2)^2/4$

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Please note that the assumption is used only in the part marked red.

Answer 2

Hint 1:

$$(k+1)^3-k^3=3k^2+3k+1 \rightarrow \sum_{k=1}^{n} [(k+1)^3-k^3]=3\sum_{k=1}^{n} k^2+3\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$$

Hint 2:

You also can use induction.

Can you finish?