# How to calculate this sum?

Let $x_1,\cdots,x_k$ be numbers between 0 and 1. Then is it possible to get explicit expression for the following sum:$$\sum_{n_1,\cdots,n_k\geq 1} x_1^{n_1}\times C_{n_1+n_2}^{n_2}\times x_2^{n_2}\times\cdots\times C_{n_{k-1}+n_k}^{n_k}\times x_k^{n_k},$$ where $$C_{m}^l=\frac{m!}{l!(m-l)!}.$$

Thanks.

This solution is based on the identity $\sum_{r=0}^\infty\,\binom{n+r}{r}x^r=(1-x)^{-n-1}$. I know that your sum starts with $n_1,n_2,\ldots,n_k\geq 1$, but for the sake of simplicity, I will calculate the sum starting with $n_1,n_2,\ldots,n_k\geq 0$. You could probably easily modify my answer to get the sum that starts with $n_1,n_2,\ldots,n_k\geq 1$.
Note that \begin{align} \sum_{n_1,\ldots,n_k\geq 0}\,\left(\prod_{r=1}^{k-1}\,\binom{n_r+n_{r+1}}{n_r}\right)\left(\prod_{j=1}^k\,x_j^{n_j}\right) &=\sum_{n_2,\ldots,n_k\geq 0}\,\left(\prod_{r=2}^{k-1}\,\binom{n_r+n_{r+1}}{n_r}\right)\left(\prod_{j=2}^k\,x_j^{n_j}\right)\left(1-x_1\right)^{-n_2-1} \\ &=\small{\textstyle\frac{1}{1-x_1}}\sum_{n_3,\ldots,n_k\geq 0}\,\left(\prod_{r=3}^{k-1}\,\binom{n_r+n_{r+1}}{n_r}\right)\left(\prod_{j=3}^k\,x_j^{n_j}\right)\textstyle\left(1-\frac{x_2}{1-x_1}\right)^{-n_3-1} \\ &=\ldots=\left(\frac{1}{1-x_1}\right)\left(\frac{1}{1-\frac{x_2}{1-x_1}}\right)\cdots\left(\frac{1}{1-\frac{x_k}{1-\frac{x_{k-1}}{\ldots\frac{1}{1-x_1}}}}\right)\,. \end{align} If $k=1$, the answer is $\frac{1}{1-x_1}$. If $k=2$, the answer is $\frac{1}{1-x_1-x_2}$. If $k=3$, the answer is $\frac{1}{1-x_1-x_2-x_3-x_1x_3}$. The answer for a general $k$ is $$\sum_{n_1,\ldots,n_k\geq 0}\,\left(\prod_{r=1}^{k-1}\,\binom{n_r+n_{r+1}}{n_r}\right)\left(\prod_{j=1}^k\,x_j^{n_j}\right)=\frac{1}{1-f_k}\,,$$ where $f_0:=0$, $f_1:=x_1$, and $f_j:=f_{j-1}+x_j+f_{j-2}\cdot x_j$ for all $j=2,3,\ldots,k$.