All power sums have a closed polynomial forms for integral powers. For example,
$1^2+2^2+3^2+\cdots+n^2=\displaystyle \sum_{k=1}^n k^2=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$
More generally
$1^m+2^m+3^m+\cdots+n^m=\displaystyle \sum_{k=1}^n k^m=\displaystyle \sum_{i=1}^{m+1} a_i n^i$
In the case of $m=2$, $a_1=\frac{1}{6}$, $a_2=\frac{1}{2}$, and $a_3=\frac{1}{3}$.
To solve for the closed form solution of every power sum we use the following process.
Take the bottom left half of Pascal's infinite matrix. Remove the top leftbottom right diagonal of 1's. Remove the top row and right most column of 0's. Invert the matrix. Now, the $j$th row from the right yields the coefficients, $a_i$ of the sum of (j1)th powers. A full proof and explanation is given in this document.
This table contains the coefficients of the closed form solutions for the first 61 power sums. As an example, the 3rd column from the right reads:
$\begin{array}{c}\\ \frac{1}{3}\\ \frac{1}{2}\\ \frac{1}{6} \end{array}$
This means $a_3=\frac{1}{3}$, $a_2=\frac{1}{2}$, and $a_1=\frac{1}{6}$. Thus the closed form solution is $a_3n^3+a_2n^2+a_1n=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n$.
I used the code below for CS50 to generate the bottom 61 rows and columns of Pascal's matrix. After eliminating the unnecessary rows and columns, I then inverted the matrix on this website (huge thanks to whomever made this). I had to fill in 455 0's (probably due to the range of a long double) by hand but don't worry, all numbers in the giant table linked above are accurate.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 

$</code> ... <code>$</code>...<code>."> Easy Math Editor
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
2 \times 3
2^{34}
a_{i1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
There are no comments in this discussion.