algebra precalculus - Power summation of $n^3$ or higher

If I want to find the formula for

$$\sum_{k=1}^n k^2$$

I would do the next: $$(n+1)^2 = a(n+1)^3 + b(n+1)^2 + c(n+1) - an^3 - bn^2 - cn$$
After calculations I will match the coefficients.
What equation should I use, if I want to calculate the coefficients of the formula that gives the sum $$\sum_{k=1}^n k^3$$ or higher?

EDIT: Please do not refer me to other places, not trying to blaim, or to be rude, but I find your sources too complicated for me, and unable to learn from, for my level.

If you can please explain how am I supposed to find an equation that matches to my needs, not the equation itself, only the technique I should be working with.

Regards, Guy

Answer

I think I know what strategy was used for the proof you are referring to for sum of squares. The same idea works for sum of cubes, but is more painful. We try to find numbers $a$, $b$, $c$, and $d$ such that

$$(n+1)^3=[a(n+1)^4 +b(n+1)^3+c(n+1)^2+d(n+1)]-[an^4+bn^3+cn^2+dn].$$
To find these numbers, there are some shortcuts. But ultimately you will probably need to compute $(n+1)^2$ (familiar), $(n+1)^3=n^3+3n^2+3n+1$, and $(n+1)^4=n^4+4n^3+6n^2+4n+1$.

We get a system of $4$ equations in $4$ unknowns, but the solution turns out to be surprisingly uncomplicated. To give a start, on the left the coefficient of $n^3$ is $1$. On the right it is $4a$, so $a=\frac{1}{4}$. As a further check when you do the work, it should turn out that $d=0$.

After you have found the remaining constants $b$ and $c$, do the "collapsing" or "telescoping" argument that you seem to have seen for $\sum_{k=1}^n k^2$.