Finite series identity

Question

How would I prove this statement? I know that it's a finite series. I don't know how to approach this at all.

$$\sum_{i=1}^N i^3 = \left(\sum_{i=1}^N i \right)^2$$

Answer 1

Do it by induction. Base case: $N=1$.

$$\sum_{i=1}^N i^3 = 1^3 = 1^2 = \left(\sum_{i=1}^N i\right)^2$$

Now let $M = N+1$. Then you have

$$\sum_{i=1}^M i^3 = \sum_{i=1}^N i^3 + M^3$$

and

$$\left(\sum_{i=1}^M i\right)^2 = \left(\sum_{i=1}^N i + M\right)^2 = \left(\sum_{i=1}^N i\right)^2 + 2M\sum_{i=1}^N i + M^2$$

Then recall the summation formula for an arithmetic series:

$$\sum_{i=1}^N i = \frac{N}{2} \cdot (1 + N) = \frac{N+N^2}{2}$$

Whereby

$$2M\sum_{i=1}^N i + M^2 = 2M\frac{N+N^2}{2} + M^2 = M(N+N^2) + M^2$$

So

$$\left(\sum_{i=1}^N i\right)^2 + 2M\sum_{i=1}^N i + M^2 = \left(\sum_{i=1}^N i\right)^2 + M(N+N^2) + M^2$$

And we want to show that this is equal to $\sum_{i=1}^N i^3 + M^3$. We know from the base case that $\sum_{i=1}^N i^3 = \left(\sum_{i=1}^N i\right)^2$. So we'll subtract that from both equations.

So now we want to show that

$$M(N+N^2) + M^2 = M^3$$

And recall $M = N+1$. So: $$M(N+N^2) = M \cdot (M-1 + (M-1)^2) = M \cdot (M-1 + M^2 - 2M + 1) $$ $$= M \cdot (M + M^2 - 2M) = M \cdot (M^2 - M) = M^3 - M^2$$

So $$M(N+N^2) + M^2 = M^3 - M^2 + M^2 = M^3$$

As desired.

Answer 2

Hint: you could try induction on $N$. It might help if you knew the formula for $\sum_{i=1}^N i$.