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?
 A: $$S(n)=(1+2n)^2=1+4n+4n^2$$
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
A: $\sum_\limits{i=0}^n 2(i + 1)^2 = 2\sum_\limits{i=1}^{n+1} i^2$
Which gets to the meat of the question, what is  $\sum_\limits{i=1}^n i^2$?
There are a few ways to do this.  I think that this one is intuitive.

In the first triangle, the sum of $i^{th}$ row equals $i^2$
The next two triangles are identical to the first but rotated 120 degrees in each direction.
Adding corresponding entries we get a triangle with $2n+1$ in every entry.  What is the $n^{th}$ triangular number?
$3\sum_\limits{i=1}^n i^2 = (2n+1)\frac {n(n+1)}{2}\\
\sum_\limits{i=1}^n i^2 = \frac {n(n+1)(2n+1)}{6}$
To find: $\sum_\limits{i=1}^{n+1} i^2 $, sub $n+1$ in for $n$ in the formula above.
$\sum_\limits{i=0}^n 2(i + 1)^2 = \frac {(n+1)(n+2)(2n+3)}{3}$
Another approach is to assume that $S_n$ can be expressed as a degree $3$ polynomial.  This should seem plausible
$S(n) = a_0 + a_1 n + a_2 n^2 + a_3n^3\\
S(n+1) = S(n) + 2(n+2)^2\\
S(n+1) - S_n = 2(n+2)^2\\
S(n+1) = a_0 + a_1 (n+1) + a_2 (n+1)^2 + a_3(n+1)^3\\
a_0 + a_1 n+a_1 + a_2 n^2 + 2a_2n+a_21 + a_3n^3 + 3a_3n^2 + 3a_3n + 1\\
S(n+1) - S(n) = (a_1 + a_2 + a_3) + (2a_2 + 3a_3) n + 3a_3 n^2 = 2n^2 + 4n + 2$
giving a system of equations:
$a_1 + a_2 + a_3 = 2\\
2a_2 + 3a_3 = 4\\
3a_3 = 1\\
a_0 = S(0)$
