What is $s_3$ and $s_4$ for $x$

Question

$\sum_{i=0}^n i^k = s_k(n)$,

$s_k$ polynomial from degree $k+1$

I have already shown for $s_2(x) = \frac{x(x+1)(2x+1)}6$

How from the sum and $s_2(x)$ can be shown for $s_3(x)$ and $s_4(x)$ respectively.

Answer 1

if $f(x)$ be the generator function of ${(a_n)}_{n \in \mathbb N}$ then $\frac{f(x)}{1-x}$ is generator function of $\sum_{i=1}^{k}a_i $.

$\frac{1}{1-x}= 1+x+ x^2+x^3 +x^4 +... $ $\rightarrow $ $\frac{x}{(1-x)^2}=x+ 2x^2+ 3x^3+4x^4+... $ $\rightarrow $ $A+\frac{x(1-x)^2-2x(1-x)}{(1-x)^4}=1+2^2x^2+3^2x^3+... $

then coefficient $x^n$ in $\frac{A}{1-x}$ is $s_2(n)$.

so $s_3(n)$ is coefficient $x^n$ in $B=\frac{xA^{\prime}}{1-x}$

and $s_4(n)$ is coefficient $x^n$ in $\frac{xB^{\prime}}{1-x}$