The sum of the fourth powers of the first $n$ positive integers

Question

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?

Answer 1

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

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

Answer 2

I hope this helps :

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

or Faulhaber's formula :

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