3
$\begingroup$

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

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

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

$\sum_{i=1}^11^3=1=\frac{1^2(2)^2}{4}$

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

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

Considering the sum on the left side:

$\sum_{i=1}^{n+1}i^3=\sum_{i=1}^ni^3+(n+1)^3$

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

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

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

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

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

  • 0
    @MichaelHardy Thanks for pointing out my misused arrow. The post has been edited to fix this.2012-10-10

3 Answers 3

4

Note that

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

  • 0
    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.2015-10-24
0

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 $n$.

0

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;

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

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

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

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

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

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