Proving that two summations are equivalent: $\sum_{i=1}^n i^3 = (\sum_{i=1}^n i)^2$

Question

Give a constructive proof to show that for all $n \geq 1$ ,

$\sum\limits_{i=1}^n i^3 = (\sum\limits_{i=1}^n i)^2$

Observe that $(n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1$ .

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Now, the two following equalities are obvious:

$\sum\limits_{i=1}^n i^3 = 1^3 + 2^3 + 3^3 + ... + n^3$

$(\sum\limits_{i=1}^n i)^2 = (1 + 2 + 3 + ... + n)^2$

And they are both obviously equivalent given the first few test cases:

$\sum\limits_{i=1}^n i^3 = A(n)$

• $A(1) = 1^3 = 1$
• $A(2) = 1^3 + 2^3 = 1 + 8 = 9$
• $A(3) = 1^3 + 2^3 + 3^3 = 9 + 27 = 36$

$(\sum\limits_{i=1}^n i)^2 = B(n)$

• $B(1) = (1)^2 = 1$
• $B(2) = (1 + 2)^2 =9 $
• $B(3) = (1 + 2 + 3)^2 = 36$

Now, I am thinking of finding the closed-forms for both functions in the hopes that they are indeed the same. Then I would prove those closed forms to work by induction. But:

1. I don't know if that would be a sound way to do it.
2. I don't know if this would even qualify as constructive, as the question requests.

As you may tell, I am no math major. I am a Computer Science major, though. This is a computing fundamentals class. I took discrete 1.5 years ago, so my knowledge is about as fresh as a litter box. I've been in quite a rut for a few hours over this.

Answer 1

Your goal is to prove the statement $S(n)$ for all $n\geq 1$ where $$ S(n) : 1^3 + 2^3 +3^3 +\cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2. $$ Using $\Sigma$-notation, we may rewrite $S(n)$ as follows: $$ S(n) : \sum_{r=1}^n r^3 = \left[\frac{n(n+1)}{2}\right]^2. $$ Base step: The statement $S(1)$ says that $(1)^3 = (1)^2$ which is true because $1=1$.

Inductive step [$S(k)\to S(k+1)$]: Fix some $k\geq 1$, where $k\in\mathbb{N}$. Assume that $$ S(k) : \sum_{r=1}^k r^3 = \left[\frac{k(k+1)}{2}\right]^2 $$ holds. To be proved is that $$ S(k+1) : \sum_{r=1}^{k+1} r^3 = \left[\frac{(k+1)((k+1)+1)}{2}\right]^2 $$ follows. Beginning with the left side of $S(k+1)$, \begin{align} \sum_{r=1}^{k+1}r^3 &= \sum_{r=1}^k r^3 + (k+1)^3\tag{evaluate sum for $i=k+1$}\\[1em] &= \left[\frac{k(k+1)}{2}\right]^2+(k+1)^3\tag{by $S(k)$}\\[1em] &= \frac{(k+1)^2}{4}[k^2+4(k+1)]\tag{factor out $\frac{(k+1)^2}{4}$}\\[1em] &= \frac{(k+1)^2}{4}[(k+2)(k+2)]\tag{factor quadratic}\\[1em] &= \frac{(k+1)^2(k+2)^2}{4}\tag{multiply and rearrange}\\[1em] &= \left[\frac{(k+1)(k+2)}{2}\right]^2\tag{rearrange}\\[1em] &= \left[\frac{(k+1)((k+1)+1)}{2}\right]^2,\tag{rearrange} \end{align} one arrives at the right side of $S(k+1)$, thereby showing that $S(k+1)$ is also true, completing the inductive step.

By mathematical induction, it is proved that for all $n\geq 1$, where $n\in\mathbb{N}$, that the statement $S(n)$ is true.

Note: The step where $\dfrac{(k+1)^2}{4}$ is factored out is an important one. If we do not factor this out and, instead, choose to expand $(k+1)^3$, the problem becomes much more messy than it needs to be.

Answer 2

We prove $\sum\limits_{i=1}^n i^3 = (\sum\limits_{i=1}^n i)^2$.

$i = 1$ means $1^3 = 1^2$.

Assume true for all $n \in \Bbb{N}$. We prove it's true for $i = n+1$.

Well, $\sum\limits_{i=1}^{n+1} i^3 = \sum\limits_{i=1}^n i^3 + (n+1)^3 = (\sum\limits_{i=1}^n i)^2 + (n+1)^3$.

Now, using the fact that $\sum\limits_{i=1}^n i = \frac{(n)(n+1)}{2}$, can you finish the proof?

Answer 3

I proposed here another way to approach: consider the identity $$ n^4=\sum_{i=1}^n {r^4-(r-1)^4}=\sum_{i=1}^n 4i^3-6i^2+4i-1.$$ It follows that $$\sum_{i=1}^n i^3=\frac{n^4+1}{4}+\frac{3}{2}\sum_{i=1}^n i^2-\sum_{i=1}^n i. $$ Simpify the expression and you get $$\sum_{i=1}^n i^3=\binom{n+1}{2}^2. $$

This method allows one recursively to find explicitely, given a positive integer $k$, the sum $\sum_{i=1}^n i^k$.