Sum of Powers of Positive Integers

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Theorem

Let $n, p \in \Z_{>0}$ be (strictly) positive integers.

Then:

\(\ds \sum_{k \mathop = 1}^n k^p\) \(=\) \(\ds 1^p + 2^p + \cdots + n^p\)
\(\ds \) \(=\) \(\ds \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}\)
\(\ds \) \(=\) \(\ds \frac {n^{p + 1} } {p + 1} + \frac {B_1 \, n^p} {1!} + \frac {B_2 \, p \, n^{p - 1} } {2!} + \frac {B_4 \, p \paren {p - 1} \paren {p - 2} n^{p - 3} } {4!} + \cdots\)

where:

$B_k$ are the Bernoulli numbers
$p^{\underline k}$ is the $k$th falling factorial of $p$.


Proof


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Sources

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