Let $a_n$ represent the summation. We have $a_n=\left(\frac{n+3}{n+1}\right)a_{n-1}+\frac{2}{n+1}$ for every $n>0$ and $a_0=0$. In fact, if the sum were $b_n:= \sum_{k=0}^{n}\ \ {\sum_{1\le i_{1}<\ldots < i_{k}\le n}^{}{\frac{2^{k}}{(i_{1}+1)(i_{2}+1)\ldots (i_{k}+1)}} }$, then $b_n=a_n+1$. We have $b_0=1$ and $b_n= \left(\frac{n+3}{n+1}\right)b_{n-1}$ for every $n>0$. Now, it is easier to show that $b_n=\frac{(n+2)(n+3)}{6}$. That is, $a_n=\frac{n(n+5)}{6}$.

Can someone please explain this solution?