Simple proof by induction: $1^3 + 2^3 +3^3 +...+ n^3 = [ (n(n+1))/2 ]^2 $

Question

I am rather illiterate when it comes to mathematics, I am afraid. In an effort to change that, I grabbed a copy of 'What is mathematics? : An elementary approach to ideas and methods' and have already encountered some difficulty. It seems silly to waste so much time trying to solve it myself, so I decided to ask for some help in solving and illuminating the taken steps, so that I can solve some more on my own. I shall review some of the basics while awaiting answers, and hope that my session tomorrow shall be more productive...

The problem is as follows:

Prove by induction that $1^3 + 2^3 +3^3 ... n^3 = [ (n(n+1))/2 ]^2 $.

As is, at the point I decided to seek help and look up material for review, I have taken the following steps:

solved for the base case: n=1, $1^3 = [2/2]^2 $

$$ 1 = 1^2, 1=1 $$

Then after proving the basis, I stated the assumption that:

$1^3 + 2^3 ... k^3 = [ (k(k+1)) / 2 ]^2$ is true.

Then I tried to solve for the next case, $(k+1)^3$:

$$ 1^3 + 2^3 + 3^3 ... k^3 + (k+1)^3 = [ ( (k(k+1)) / 2 ) + (k+1) ]^2 $$

$$ [ (k(k+1)) / 2 ]^2 + (k+1)^3 = [ (k(k+1) + 1(k+1) ) / 2 ]^2 $$

at this point after failing for a while and having spent quite some time looking up material, I sought help. I hope I'm going down the right path here, but I suspect I shall soon find out...

Help is greatly appreciated.

Answer 1

**HINT:**$$\left(\frac{k(k+1)}{2}\right)^2+(k+1)^3=\frac{k^2(k+1)^2}{4}+(k+1)^3$$$$=\frac{k^2(k+1)^2+4(k+1)^3}{4}$$$$=\frac{(k+1)^2(k^2+4(k+1))}{4}$$$$=\frac{(k+1)^2(k^2+4k+4)}{4}$$$$=\frac{(k+1)^2(k+2)^2}{4}$$Hopefully you can complete the proof from here...

Answer 2

You're on the right track! The simplest way to show the last line is to foil out the terms on the left and right and note that they're equal (tedious, I know--but it'll get you what you need). If you haven't done so already, try proving these formulas with induction (for extra practice; they're a little easier): $$\sum_{k=1}^n k = \frac{n(n+1)}{2}\quad\quad \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}.$$

EDIT: And by "last line," I mean the "goal": $$\left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2.$$

Answer 3

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.