Sum of a series problem involving cubes For any odd integer $n$, evaluate $n^3 - (n-1)^3 + \cdots + (-1)^{n-1} \cdot 1^3 = \sum_{k=1}^n (-1)^{k+1}\cdot k^3$
How would you go about solving such a problem?
Any help would be appreciated.

I would use the following ideas:
(1): Standard formula: $1^3+2^3+\cdots+a^3=\left(\frac{a(a+1)}{2}\right)^2$
(2): Use (1) to get an expression for $2^3+4^3+6^3+\cdots+(2a)^3=2^3\cdot(1^3+2^3+\cdots+a^3)$
(3): Use (1) and (2) to get an expression for $1^3+3^3+5^3+\cdots+(2a-1)^3$ by noting that
$$1^3+3^3+5^3+\cdots+(2a-1)^3=\left(1^3+2^3+\cdots+(2a)^3\right)-\left(2^3+4^3+\cdots+(2a)^3 \right)$$
(4) Your expression breaks down to an expression from (3) minus an expression from (2) (since your $n$ is supposed to be odd--if your $n$ were even you could still use these ideas to get your expression for the alternating sum.)
