calculus - Partial sum of divergent series

I am trying to find the nth partial sum of this series:
$S(n) = 2(n+1)^2$

I found the answer on WolframAlpha:

$\sum_{n=0}^m (1+2n)^2 =\frac{1}{3}(m+1)(2m+1)(2m+3)$

How can I calculate that sum, without any software?

Answer

$$S(n)=(1+2n)^2=1+4n+4n^2$$
You can now use the following $$\sum_{n=0}^m1=m+1\\\sum_{n=0}^mn=\frac{m(m+1)}{2}\\\sum_{n=0}^mn^2=\frac{m(m+1)(2m+1)}{6}$$

Alternatively, compute the first 4-5 elements. The sum of a polynomial of order $p$ will be a polynomial of order $p+1$ in the number of terms. Find the coefficients, then prove by induction