Alternating Sum of Cubes

Question

How is it possible to evaluate:

$$\sum_{k=1}^n{((-1)^{n-k}\cdot k^3)}=n^3 - (n-1)^3 + (n - 2)^3 - \cdots \pm 1^3$$

The fact that there is the $\pm$ at the end makes it difficult.

Answer 1

No, the $\pm$ at the end does not make it difficult at all. Just split the problem up into "odd" and "even" cases. (See Andre Nicolas's answer for detail.)

Alternatively, notice that if we define:

$$A_n = \sum_{k=1}^n{((-1)^{n-k} \cdot k^3)}$$

for the terms in a sequence, then:

$$A_n + A_{n+1} = (n+1)^3$$

It is possible to be "fancy" and solve for the formula by using generating functions!

The answer is $$\frac{1}{2}n^3 + \frac{3}{4}n^2-\frac{1}{8}+\frac{1}{8}(-1)^n$$ by the way. See the OEIS sequence ($1$, $7$, $20$, $44$, $81$, $\cdots$), for example.

EDIT: If you're interested, the generating functions way goes kind of like this: Define $f(x)$ to be:

$$f(x) = \sum_{k=0}^\infty{(A_kx^k)}$$

Then, calculate $(x+1)f(x)=xf(x)-f(x)$ (as a rational function) by applying:

$$A_n + A_{n+1} = (n+1)^3$$

This gives the generating function of $f(x)$ (it is a rational function). Expand it using partial fractions; each partial fraction is a geometric series or a power (i.e., derivative) of a geometric series.

Express each partial fraction as a simple series, add them, and equate coefficients to $A_n$. And you get the answer!!!

Answer 2

Outline: It is useful to distinguish between $n$ odd and $n$ even.

For $n$ odd our sum is $$\left(n^3+(n-1)^3+\cdots +1^3\right)-2\left((n-1)^3+(n-3)^3+\cdots +2^3\right).$$ You know a closed form for the sum of the first $n$ cubes. And $(n-1)^3+(n-3)^3+\cdots +2^3$ is $8$ times the sum of the first $\frac{n-1}{2}$ cubes.

A similar trick works for even $n$.

Alternately, note that $$k^3-(k-1)^3=3k^2-3k+1.$$ Group in pairs. We know how to sum $k^2$, $k$, and $1$.