Quick Clarification Quite Curious

[bomba923]

Quite Curious

$$\begin{gathered} \sum\limits_{i = 1}^n i = \frac{{n\left( {n + 1} \right)}} {2} \hfill \\ \sum\limits_{i = 1}^n {i^2 } = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}} {6} \hfill \\ \sum\limits_{i = 1}^n {i^3 } = \frac{{n^2 \left( {n + 1} \right)^2 }} {4} \hfill \\ \vdots \hfill \\ \left( {etc} \right) \hfill \\ \end{gathered}$$
------------------------------------------------
But in general,

$$\forall k \in \mathbb{N} ,$$

what is the general summation formula for

$$\sum\limits_{i = 1}^n {i^k } \; {?}$$

 

[shmoe]

You can get something in general in terms of Bernoulli numbers:

http://en.wikipedia.org/wiki/Bernoulli_number

Mathworld gives it explicitly up to the sum of 10th powers:

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

 

[tacman]

The general summation formula for $\sum\limits_{i = 1}^n {i^k }$ is $\frac{n^{k+1}}{k+1}$. This formula can be derived using the method of finite differences or by using the binomial theorem. It is a useful formula for finding the sum of a series of numbers raised to a power, and can be applied to various mathematical and scientific problems.