The sum of the fourth powers of the first $n$ positive integers 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
$$a=\frac15,b=\frac12,c=\frac13,d=0,e=-\frac1{30}.$$
