Product of indexed parentheses

Question

I am struggling to see whether is it possible to find a closed form expression for the following product

$$(1+K)(1+2K)(1+3K)...(1+(n-1)K)$$

where $K$ is a constant. I tried to come up with a factorial type of answer but I could not.

Any ideas?

Answer 1

This is $$K^{n-1}\left(1+\frac1K\right)\left(2+\frac1K\right)\cdots \left(n-1+\frac1K\right).$$ Using $\Gamma(x+1)=x\Gamma(x)$ gives $$K^{n-1}\frac{\Gamma(n+1/K)}{\Gamma(1+1/K)}.$$

Answer 2

Some numbers if you want to do some pattern searching ...

Answer 3

I don't know if this is what you want but if the expression above was $f(n,K)$, the following is true.

$f(n,K)=\sum_{j=0}^{n}s(n,n-j)K^j$ where $s(n,k)$ are the Stirling numbers of first kind.