Need to evaluate the series $\sum_{i = 0}^n i^5$

Question

I need help in evaluating the following sum:

$$\sum_{i = 0}^n i^5 $$

I can evaluate series when they are arithmetic or geometric but I don't know how to solve this one.

Answer 1

Main Idea: Knowing the closed form of $\sum\limits_{k=1}^n k^{1},\ldots,\sum\limits_{k=1}^n k^{r}$ let's you determine the closed form of $\sum\limits_{k=1}^n k^{r+1}$.

The Procedure: We begin with the formula :

$$(k+1)^6-k^6=6 k^5+15 k^4+20 k^3+15 k^2+6 k+1.$$

Writing it for $k=1,\ldots,n$ and adding we obtain :

$$\begin{array}{rcl} 2^6-1^6 &=& 6\cdot 1^5+15\cdot 1^4+20\cdot 1^3+15\cdot 1^2+6\cdot 1+1 \\ 3^6-2^6 &=& 6\cdot 2^5+15\cdot 2^4+20\cdot 2^3+15\cdot 2^2+6\cdot 2+1 \\ \cdot &&\\ \cdot &&\\ \cdot &&\\ (n+1)^6-n^6&=&6\cdot n^5+15\cdot n^4+20\cdot n^3+15\cdot n^2+6\cdot n+1 \\ \hline (n+1)^6-1&=&6[1^5+\cdots+n^5]+15[1^4+\cdots+n^4]+20[1^3+\cdots+n^3]\\&&+15[1^2+\cdots+n^2]+6[1+\cdots+n]+[\underbrace{1+\cdots+1}_{n\,\rm times}]. \end{array}$$

Knowing that

$$\sum_{k=1}^n1=n,\qquad \sum_{k=1}^n k=\frac{n(n+1)}{2},\qquad \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6},\qquad\\ \sum_{k=1}^nk^3=\dfrac{n^2(n+1)^2}{4},\qquad\sum_{k=1}^nk^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30},$$

you can solve for $1^5+\cdots+n^5$, although that is a tedious task.

Answer 2

HINT:

Use $(r+1)^{m+1}-r^{m+1}=\binom{m+1}1r^m+\binom{m+1}2r^{m-1}+\cdots+\binom{m+1}{m-1}r^2+\binom{m+1}mr+1$

To know $\sum_{r=1}^nr^u,$ set $m=u$

Assumption : $\sum_{r=1}^nr^v,0\le v\le u-1$ are known

Answer 3

Take a look at Faulhaber's formula on wikipedia.

Note that even though finding the result is difficult, once it has been found is pretty easy to show that it's correct using induction

To answer your question,

$$\sum_{i=0}^n i^5 = \frac{n^2(n+1)^2(2n^2+2n-1)}{12}$$