bernoulli

Closed forms of the sum of powers for fixed values of m

S_m(n) = \sum_{k=1}^n k^m = 1^m + 2^m + \cdots + n^m \,

are always polynomials in n of degree m + 1. Note that Sm(0) = 0 for all m ≥ 0 because in this case the sum is the empty sum. The coefficients of these polynomials are related to the Bernoulli numbers by Bernoulli's formula:

S_m(n) = {1\over{m+1}}\sum_{k=0}^m {m+1\choose{k}} B_k n^{m+1-k}

Let n ≥ 0. Taking m to be 0 and B0 = 1 gives the natural numbers 0, 1, 2, 3, … (sequence A001477).

0 + 1 + 1 + \cdots + 1 = \frac{1}{1}\left(B_0 n\right) = n.

Taking m to be 1 and B1 = 1/2 gives the triangular numbers 0, 1, 3, 6, … (sequence A000217).

0 + 1 + 2 + \cdots + n = \frac{1}{2}\left(B_0 n^2+2B_1 n^1\right) = \frac{1}{2}\left(n^2+n\right).

Taking m to be 2 and B2 = 1/6 gives the square pyramidal numbers 0, 1, 5, 14, … (sequence A000330).

0 + 1^2 + 2^2 + \cdots + n^2 = \frac{1}{3}\left(B_0 n^3+3B_1 n^2+3B_2 n^1 \right) = \frac{1}{3}\left(n^3+\frac{3}{2}n^2+\frac{1}{2}n\right).

Although Bernoulli's formula reflects faithfully what Bernoulli has written some authors state Bernoulli's formula in a different way which is neither in accordance with Bernoulli's statement nor has any obvious advantage compared to it. They write:

S_m(n) = {1\over{m+1}}\sum_{k=0}^m (-1)^k {m+1\choose{k}} B_k n^{m+1-k}