Summation Identity for i^p power question, really simple

[Asphyxiated]

Homework Statement

\sum_{i=0}^{n} i^{p} = \frac {(n+1)^{p+1}}{p+1} + \sum_{k=1}^{p} \frac {B_{k}}{p-k+1} (^{p}_{k}) (n+1)^{p-k+1}


where B_k is a Bernoulli number.

There is no actual question here I would just like to know if this formula is for sums of i to any power, of course its rather cumbersome but the question still stands. All I want to know is if I understand what it is doing.

edit: oh and

(^{p}_{k}) = \frac {p!}{k!(p-k)!} \;\; 0 \leq k \leq p

 

[Gib Z]

Yes, that is called Faulhaber's formula and is valid for natural numbers p.

 

[ystael]

Incidentally, there is a better way to make binomial coefficients than (^p_k), which will always produce "small" binomial coefficients and won't always place the arguments gracefully with respect to the parentheses. You can write \binom{p}{k} to get \binom{p}{k}.