Prove the following statement $S(n)$ for $n\ge1$:


To prove the basis, I substitute $1$ for $n$ in $S(n)$:


Great. For the inductive step, I assume $S(n)$ to be true and prove $S(n+1)$:


Considering the sum on the left side:


I make use of $S(n)$ by substituting its right side for $\sum_{i=1}^ni^3$:


This is where I get a little lost. I think I expand the equation to be


but I'm not totally confident about that. Can anyone provide some guidance?

  • $\begingroup$ You are on the right path, just expand $(n+1)^2(n+2)^2$ and match the coefficients with your last formula (wherein you can also expand $(n+1)^4$). $\endgroup$
    – filmor
    Oct 10, 2012 at 15:04
  • 5
    $\begingroup$ Some inductive proofs can be found here. In particular this answer is along the same lines as your attempt. $\endgroup$ Oct 10, 2012 at 15:05
  • $\begingroup$ You're making one of the usual mistakes: you write $\sum_{i=1}^11^3=\frac{1^2(2)^2}{4}=>1=1$. You have the arrow ($=>$) presumably meaning "if...then..." going in the wrong direction. What you need is $\sum_{i=1}^1 i^3 = 1 = \dfrac{1^2\cdot2^2}{4}$. That shows that the first expression in this string of equalities equals the third expression. This is a flaw in logical presentation that it's notoriously hard to talk students out of. $\endgroup$ Oct 10, 2012 at 16:00
  • $\begingroup$ ....but that doesn't upset your later argument. At the end, you're doing OK; you just need to keep going. The common denominator is $4$, so you need $\dfrac{4(n+1)^3}{4}$ $=\dfrac{4(n^3+3n^2+3n+1)}{4}$ $\dfrac{4n^3+12n^2+12n+4}{4}$, and then add the two fractions and simplify, and then factor the numerator. $\endgroup$ Oct 10, 2012 at 16:06
  • $\begingroup$ @MichaelHardy Thanks for pointing out my misused arrow. The post has been edited to fix this. $\endgroup$
    – Brian
    Oct 10, 2012 at 19:30

3 Answers 3


Note that

$$\frac{n^{2}(n+1)^{2}}{4}+(n+1)^{3} =\frac{(n+1)^{2}}{4}[n^{2}+4(n+1)] =\frac{(n+1)^{2}(n+2)^{2}}{4}$$

  • $\begingroup$ +1 I don't know why so many people want to expand the $(n+1)^3$ when this approach is far clearer. $\endgroup$ Oct 10, 2012 at 19:42
  • $\begingroup$ Especially when you want the final form to have an $(n+1)^2$ factor. Factoring it out from the start just makes sense. $\endgroup$
    – Mike
    Oct 10, 2012 at 20:04
  • $\begingroup$ My algebra is rusty and the intellectual leap to $\frac{(n+1)^2}{4}[n^2+4(n+1)]$ in the first equality escapes me. Can you explain further? $\endgroup$
    – Brian
    Oct 10, 2012 at 20:39
  • $\begingroup$ As you are going to be factoring common terms and as expanding and then factoring is hard. If there are common terms to begin with factor them out from the get go. Both of the sum terms have $(n+1)^2$ so just factor that out from the start. It's much easier than to expand it first and then refactoring later. Particularly as you know what you want the answer to be. $\endgroup$
    – fleablood
    Oct 24, 2015 at 6:43

So far you are doing fine. Now expand $(n+1)^3$ to a polynomial, combine it with the $\frac14 (n^4+2n^3+n^2)$ term, and check to see if the result is what you expect. What you expect should be $n^2(n+1)^2\over 4$, except with $n+1$ in place of $n$.


Here's yet another way:

You have $\frac{n^2(n+1)^2}{4}+(n+1)^3$ and you want $\frac{(n+1)^2(n+2)^2}{4}$

So manipulate it to get there;

$\frac{n^2(n+1)^2}{4}+(n+1)^3 =$

$\frac{(n^2 + 4n + 4)(n+1)^2 - (4n + 4)(n+1)^2}{4}+(n+1)^3 =$

$\frac{(n+2)^2(n+1)^2}{4}- \frac{ (4n + 4)(n+1)^2}{4}+(n+1)^3 =$

$\frac{(n+2)^2(n+1)^2}{4}-(n + 1)(n+1)^2+(n+1)^3 =$

$\frac{(n+2)^2(n+1)^2}{4}-(n+1)^3+(n+1)^3 =$



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