Trying to prove $\sum_{i=1}^{N} i^3 = (\sum_{i=1}^{N} i)^2$

Question

I'm trying to prove $\sum_{i=1}^{N} i^3 = (\sum_{i=1}^{N} i)^2$ but I got stuck along the way. This is what I have so far:

The base case is true when $N =1$.

Then for the inductive step I did:

Assume $\sum_{i=1}^{N} i^3 = (\sum_{i=1}^{N} i)^2$ is true for $1 \leq k \leq N$.

Prove $\sum_{i=1}^{N+1} i^3 = (\sum_{i=1}^{N+1} i)^2$.

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

$\hspace{25pt}= (\sum_{i=1}^{N} i^3) + (N+1)^3$

$\hspace{25pt} = (\sum_{i=1}^{N} i)^2 + (N+1)^3$ (by induction)

$\hspace{25pt} = \hspace{5pt}?$

From there I'm not sure how to do the proper conversions into equation/sigma notation to prove the right side. Can anyone help me on how to do this? Thanks.

Answer 1

\begin{align} \\ k = N \\ Assume \sum_{i=1}^{N} i^3 = (\sum_{i=1}^{N} i)^2 \\ \\ k = N+1 \\ LHS = \sum_{i=1}^{N+1} i^3 \\ = (\sum_{i=1}^{N} i^3) + (N+1)^3 \\ \\ RHS = (\sum_{i=1}^{N+1} i)^2 \\ = [\sum_{i=1}^{N} i +(N+1)]^2 \\ = (\sum_{i=1}^{N} i)^2+2(\sum_{i=1}^{N} i)(N+1)+(N+1)^2 \\ = (\sum_{i=1}^{N} i)^2+2({N(N+1) \over 2})(N+1)+(N+1)^2 \\ = (\sum_{i=1}^{N} i)^2+N(N+1)^2+(N+1)^2 \\ = (\sum_{i=1}^{N} i)^2+(N+1)^2(N+1) \\ = (\sum_{i=1}^{N} i)^2+(N+1)^3 \\ = LHS \end{align}

Prove by induction.

Answer 2

So you wish to prove $$\left(\sum_{i=1}^{N+1} i\right)^2 = \left(\sum_{i=1}^N i\right)^2 + (N+1)^3 $$

Letting $a = \sum_{i=1}^N i$ and $b = N+1$, using the formula $(a+b)^2 = a^2+2ab+b^2$, we obtain

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

It suffices to prove $(N+1)^3 = 2(N+1)\left(\sum_{i=1}^N i\right) + (N+1)^2$. By high-school mathematics we know $$\sum_{i=1}^N i = \frac{1}{2}N(N+1)$$ and therefore $$2(N+1)\left(\sum_{i=1}^N i\right) + (N+1)^2 = N(N+1)^2+(N+1)^2=(N+1)^3$$

Answer 3

You're on the right track . You can use the well-known sum : $$\sum_{i=1}^{N} i=\frac{N(N+1)}{2}$$ and then plug it in what you wrote : $$\sum_{i=1}^{N+1}i^3=\frac{N^2(N+1)^2}{4}+(N+1)^3$$

Now if you could prove that this is exactly $\frac{(N+1)^2(N+2)^2}{4}$ you'd be done because the latter is precisely $$\left (\sum_{i=1}^{N+1}i \right)^2$$

You can do this by expansion but it's easier with factoring : $$\frac{(N+1)^2(N+2)^2}{4}-\frac{N^2(N+1)^2}{4}=\frac{(N+1)^2}{4} \cdot ((N+2)^2-N^2)=\frac{(N+1)^2}{4} \cdot (4N+4)=(N+1)^3$$ as wanted .

So it follows by induction that $$\sum_{i=1}^{N} i^3=\left (\sum_{i=1}^{N} i \right )^2$$ for every $N$ .

Answer 4

The shortest,in my opinion, is to use finite differences:

Set $\;A_N=\displaystyle\sum_{i=1}^N i^3\;$ and $\;B_N=\displaystyle\sum_{i=1}^N i$. To prove $A_N=B_N^2$, we can prove

i) $A_1=B_1^2$, which is trivial.

ii) $\Delta A_N=A_{N+1}-A_N=\Delta B_N^2$ for all $N$.

Indeed, $\Delta A_N=(N+1)^3$. On the other hand: \begin{align*} \Delta B_N^2&= B_{N+1}^2 - B_N^2=(B_{N+1} - B_N)(B_{N+1} + B_N)\\ &=(N+1)\biggl(\frac{(N+1)(N+2)}2+\frac{N(N+1)}2\biggr)\\ &=\frac{(N+1)^2}2(2N+2)=(N+1)^3. \end{align*} Ο.Ε.Δ.