Summation of series Write down the sum of $\displaystyle \sum_1^{2N} n^3$ in terms of $N$, and hence find:
$1^3 - 2^3 + 3^3 - 4^3 + \cdots - (2N)^3$ in terms of $N$, simplifying your answer.
I found this to be $n^2(2n+1)^2$ but the next part is not making sense to me.
Why is the general term of this sum $-(2N)^3$, where it doesn't work for N=0 etc?
thanks

HINT:
As
$$
\begin{align}
& \sum_{1\le r\le n}r^3=\frac{n^2(n+1)^2}4 \\[10pt]
& 1^3 - 2^3 + 3^3 - 4^3 + \cdots - (2N)^3 \\[10pt]
& =\sum_{1\le r\le 2N}r^3-2\sum_{1\le r\le N}(2r)^3 \\[10pt]
& =\sum_{1\le r\le 2N}r^3-2\cdot8\sum_{1\le r\le N}r^3 \\[10pt]
& =\frac{(2N)^2(2N+1)^2}4-16\frac{N^2(N+1)^2}4 \\[10pt]
& =N^2\{(2N+1)^2-4(N+1)^2\} \\[10pt]
& =-N^2(4N+3)
\end{align}
$$
