I am solving a following problem.
Let's have $(x^1+\ldots+x^k)^n$, then find a formula for coefficients $a_r$. ($a_r$ is a coefficient at $x^r$.)
I used the formula for geometric series to get $x^n\left(\frac{1-x^k}{1-x}\right)^n$. Then using the generalized binomial theorem I got $x^n\cdot\left(\sum_{i=0}^{\infty}\binom{n+i-1}{n-1}x^i\right)\cdot\left(\sum_{i=0}^n\binom{n}{i}(-x)^{ik}\right)$. From this, I managed to extract a formula for $a_r$ in a form
$a_r=\sum_{i=0}^{\lfloor\frac{r-n}{k}\rfloor}(-1)^i\binom{n}{i}\binom{r-ik-1}{n-1}$
Does there exist a shorter formula? E.g. without summation or with fewer terms.