How can I solve a complicated induction exercise? (formula for sum of fourth powers) I have an induction exercise:
It is for $n \in \mathbb{N_0}$. Show:
$$ \sum\limits_{k=1}^{n} k^4 = \frac 1 {30} n(n+1)(2n+1)(3n^2+3n-1) $$
As far as I understand it, you have to put in $(n+1)^4$ at the end and you have to resolve it to the old function replacing $n$ with $n+1$. So I have on my paper:
$$ \frac 1 {30} (n+1)((n+1)+1)(2(n+1)+1)(3(n+1)^2+3(n+1)-1) = \frac 1 {30} n(n+1)(2n+1)(3n^2+3n-1) + (n+1)^4 $$
How can I solve this exercise? I probably have to disprove it because it is not true for numbers $> 1$.
All the other examples were with very tiny formulas.
I really tried to find a solution; what should I do?

By induction we show that:
$$1+2^4+3^4+....+n^4 =\frac 1 {5}n^5+\frac 1 {2}n^4+\frac 1 {3}n^3-\frac 1 {30}n$$
For $n=1$:
$$1=\frac 1 {30}\cdot2\cdot3\cdot5=1$$
Now suppose that is true for $n$, and show that and show that it is also true for $n+1$, that is:
$$1+2^4+3^4+....+n^4+(n+1)^4 =\frac 1 {5}(n+1)^5+\frac 1 {2}(n+1)^4+\frac 1 {3}(n+1)^3-\frac 1 {30}(n+1)$$
This equality is verified if the following equality is verified:
$$\frac 1 {5}(n+1)^5+\frac 1 {2}(n+1)^4+\frac 1 {3}(n+1)^3-\frac 1 {30}(n+1)=$$
$$=\frac 1 {5}n^5+\frac 1 {2}n^4+\frac 1 {3}n^3-\frac 1 {30}n + (n+1)^4$$
, calculating the solution with a few additions!
