# Partial sum for series, sum of cubes

1. Nov 23, 2012

### Dobsn

1. The problem statement, all variables and given/known data

I have this series

$1^{3}-2^{3}+3^{3}-4^{3}+5^{3}-6^{3} + \ldots$

2. Relevant equations

and sequence of partial sums for this series that is:

$S_n = \sum_{k=0}^{n}(-1)^{k+1} k^3 = \dfrac{1 + (-1)^n(4n^3 + 6n^2-1)}8 =\begin{cases} \dfrac{2n^3+3n^2}4; & n \text{ is even}\\ \dfrac{1-3n^2-2n^3}4; & n \text{ is odd} \end{cases}$

What I need to are finding the steps to this partial sums formula

3. The attempt at a solution

I've tried by finding partial sums for even and odd cubes and then substracting them, but it gives wrong solution.
Odd: $n^2(2n^2-1)$
Even: $2n^2(n+1)^2$

Any tip appreaciated. :)

2. Nov 23, 2012

### haruspex

3. Nov 24, 2012

### Dobsn

These are partial sums for even and odd numbers already:

even:
$S_n =\sum\limits_{k=0}^n {{2n^2(n+1)^2}}$

odd:
$S_n =\sum\limits_{k=0}^n {n^2(2n^2-1)}$

and substracting even from odd partial sums gets me:

$S_n =\sum\limits_{k=0}^n {-4^3-3n^2}$

And that doesn't get me to:
$S_n = \sum_{k=0}^{n}(-1)^{k+1} k^3 = \dfrac{1 + (-1)^n(4n^3 + 6n^2-1)}8 =\begin{cases} \dfrac{2n^3+3n^2}4; & n \text{ is even}\\ \dfrac{1-3n^2-2n^3}4; & n \text{ is odd} \end{cases}$

4. Nov 24, 2012

### Millennial

Why don't you make use of a functional equation for both the even and the odd sums, then do the subtraction? For example, you can say $f(0)=0$, $f(x+1)=f(x)+(2x+1)^3$ for the sum of the cubes of the first x odd naturals. Then, you can solve that functional equation (I hope you know how to do that?) Do the same for the even naturals, then subtract.

5. Nov 24, 2012

### haruspex

Presumably you meant
$S_n =\sum\limits_{k=0}^n {-4n^3-3n^2}$
I assume you intended to take the difference of two consecutive sums, one odd, one even. But you appear to have taken differences using the same n. You need to use, say, 2n and 2n+1. And to complete the inductive step you will need to do 2n-1 to 2n as well.