Finding closed form for $1^3+3^3+5^3+...+n^3$ I'd like to find a closed form for $1^3+3^3+5^3+...+n^3$ where $n$ is an odd number.
How would I go about doing this?
I am aware that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$ but I'm not too sure how to proceed from here. 
My gut feeling is telling me to multiply the above series by 8, then subtract it from the original, but it doesn't quite get me there because I'm going to have a whole lot of extra terms that I do not want. 

Using repeated differences we get
$$
\begin{array}{llll}
1 & 28 & 153 & 496 & 1225 & 2556 & \\
27 & 125 & 343 & 729 & 1331 & \\
98 & 218 & 386 & 602 & \\
120 & 168 & 216 & \\
48 & 48 & \\
0 & \\
\end{array}
$$
Newton's interpolation formula then gives us
$$
1 \binom{n-1}{0} + 27 \binom{n-1}{1} + 98 \binom{n-1}{2} + 120 \binom{n-1}{3} + 48\binom{n-1}{4}
=
2 n^4-n^2
$$
This is a technique well worth knowing. (The simplification was done with WA.)
