For the given power sum $ \left( \displaystyle \sum_{k=0}^{K-1} x^k \right)^n$, what is the coefficient of $x^{m}$ term?

Question

For the given power sum $ \left( \displaystyle \sum_{k=0}^{K-1} x^k \right)^n$, what is the coefficient of $x^{m}$ term ?

Let

$ \left( \displaystyle \sum_{k=0}^{K-1} x^k \right)^n = \displaystyle \sum_{m}^{} a_m x^m $

Then $a_m=?$

Answer 1

In terms of Iverson brackets and falling Pochhammer symbols,$$a_m=[x^m]\left(1-x^K\right)^n(1-x)^{-n}\\=\sum_{a,\,b\ge0}(-1)^{a+b}\binom{n}{a}\frac{(-n)_b}{b!}[Ka+b=m]\\=n\sum_{a=0}^{\max\left\{\left\lfloor\frac{n}{K}\right\rfloor,\,m\right\}}(-1)^a\frac{(n+m-Ka-1)!}{a!(n-a)!(m-Ka)!}.$$I'm not sure we can get a nicer expression than that for the integer $a_m$, except when $K\in\{0,\,1\}\lor n=0$.

Answer 2

Here's a start. I may do more later.

$\begin{array}\\ s_m(x) &=\left( \displaystyle \sum_{k=0}^{m-1} x^k \right)^n\\ &=\left(\dfrac{1-x^m}{1-x} \right)^n\\ &=\dfrac{(1-x^m)^n}{(1-x)^n}\\ &=(1-x^m)^n\sum_{i=0}^{\infty} \binom{n+i-1}{j}x^i\\ &=\sum_{j=0}^n\binom{n}{j}(-1)^{n-j}x^{m(n-j)}\sum_{i=0}^{\infty} \binom{n+i-1}{j}x^i\\ &=\sum_{j=0}^n\binom{n}{j}(-1)^{j}x^{mj}\sum_{i=0}^{\infty} \binom{n+i-1}{i}x^i\\ &=\sum_{j=0}^n\binom{n}{j}(-1)^{j}x^{mj}\sum_{i=0}^{\infty} \binom{n+i-1}{i}x^i\\ \end{array} $