Is there a way to determine what the sum of x to the power m (from 1 to n) evaluates to?

Question

I know that: $\sum_{x=1}^n x= n(n+1)/2$ but is there a way to generalize that to any power of $x$?

In other words, is there a way to determine what $\sum_{x=1}^n x^m$ is equal to for any $m\in\mathbb{N}$? Is there a clean formula for values where $x\lt1$?

I know I can find the sum of squares and cubes pretty easily but I'm having a hard time finding a general formula.

Answer 1

Faulhaber's formula gives the sum of the $m$th powers of the first $n$ natural numbers.

Answer 2

The general formula uses something called Bernoulli number

http://mathworld.wolfram.com/PowerSum.html

https://people.reed.edu/~jerry/311/bernoulli.pdf

Answer 3

As others have pointed out, the final formula is related to Bernoulli numbers. But if are not familiar with them, you can still find these formulas with a bit of calculation recursively.

Let's denote $S_n^m = \sum_{k=1}^n k^m$, with $S^0_n := n$. The basic idea is the identity $$ (k+1)^{m+1}-k^{m+1} = \sum_{\ell=0}^{m} \binom{m+1}{\ell} k^{\ell} $$ By doing a sum over $k$, we find $$ (n+1)^{m+1}-1 = \sum_{\ell=0}^{m} \binom{m+1}{\ell} S^\ell_n $$

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For example,

• $m=1$: Since $S^0_n=n$ $$ (n+1)^2-1 = n + 2S_n^1\Longrightarrow \boxed{S_n^1= \frac{n(n+1)}{2}} $$
• $m=2$, $$ (n+1)^3-1 = n + 3S_{n}^1 + 3S_{n}^2\Longrightarrow \boxed{S_n^2 = \frac{n(n+1)(2n+1)}{6}} $$ and so on...