Find a closed form for $\sum_{k=0}^{n} k^3$ Find a closed form for $\sum_{k=0}^{n} k^3$.
I would appreciate ideas for approaching questions like this in general as well.
Thanks.

First, we know that: $$\sum_{i \mathop = 1}^n i = \frac {n \left({n + 1}\right)} 2.$$ Thus: $$\left({\sum_{i \mathop = 1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4.$$ Next we use induction on $n$ to show that $$\sum_{i \mathop = 1}^n i^3 = \frac{n^2 \left({n + 1}\right)^2} 4.$$
The base case holds since $$1^3 = \frac{1 \left({1 + 1}\right)^2} 4.$$
Now we need to show that if it holds for $n$, then it holds for $n+1$. $$\eqalign{
\sum_{i \mathop = 1}^n i^3 + \left({n + 1}\right)^3 &= \sum_{i \mathop = 1}^n i^3 + \left({n + 1}\right)^3 \\
&= \frac{n^2 \left({n + 1}\right)^2} 4 + \left({n + 1}\right)^3 \\
&= \frac{n^4 + 2 n^3 + n^2} 4 + \frac {4 n^3 + 12 n^2 + 12 n + 4} 4 \\
&=  \frac{n^4 + 6 n^3 + 13 n^2 + 12 n + 4} 4 \\
&=  \frac{\left({n + 1}\right)^2 \left({n + 2}\right)^2} 4.
}$$
So by induction we have that: $$\boxed{\ \displaystyle\sum_{i \mathop = 1}^n i^3 = \frac{n^2 \left({n + 1}\right)^2} 4. \ }$$
