# The sum of the fourth powers of the first $n$ positive integers [duplicate]

I am studying mathematical induction and most of the times I have to prove something. Like, for example:

$1 + 4 + 9 + ...+ n^2 = \frac{n(n+1)(2n+1)}{6}$

This time I found a question that ask me to find a formula for

$1 + 16 + 81 + .... + n^4$

How can I do this with induction? And is there really a formula for this sum?

As $S_0=0$ and $S_n-S_{n-1}=n^4$, $S_n$ must be a polynomial of the fifth degree with no independent term, let

$$S_n=an^5+bn^4+cn^3+dn^2+en.$$

Then

$$S_n-S_{n-1}=\\ a(n^5-n^5+5n^4-10n^3+10n^2-5n+1)+ \\b(n^4-n^4+4n^3-6n^2+4n-1)+\\ c(n^3-n^3+3n^2-3n+1)+\\ d(n^2-n^2+2n-1)+\\ e(n-n+1)=\\ a(5n^4-10n^3+10n^2-5n+1)+ \\b(4n^3-6n^2+4n-1)+\\ c(3n^2-3n+1)+\\ d(2n-1)+\\ e.$$

By identification with $n^4$,

$$\begin{cases}5a=1\\-10a+4b=0\\10a-6b+3c=0\\-5a+4b-3c+2d=0\\a-b+c-d+e=0.\end{cases}$$

This is a triangular system, which readily gives

$$a=\frac15,b=\frac12,c=\frac13,d=0,e=-\frac1{30}.$$

• I like this, because it is general, no need to know the formula in advance, you will figure it out with this method. Though it doesn't give the usual "factorized" presentation right away. – zwim Dec 20 '16 at 21:02

I hope this helps :

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

or Faulhaber's formula :

https://en.wikipedia.org/wiki/Faulhaber%27s_formula