How to find if this summation equals zero? For every value of $n_1$ i tested with, it came out as $0$. However I have no idea how to prove if it does or not for every value of ${n_1}$.
Note that ${n_1}$ has to be smaller than ${n_0}$ in the summation.
$$\sum_{i=1}^{n_1} \frac{(-1)^{i+1}2^{{n_1}-i}i({n_0}({n_1}-2+i)+{n_1}(i-1)+i^2-i)}{i!({n_1}-i)!({n_0}+i)({n_0}+i-1)({n_0}+{n_1})}$$
For instance, plugging $2$ to ${n_1}$ and solving the summation gives $\frac{2{n_0}}{({n_0}+1){n_0}({n_0}+2)}-\frac{2(2{n_0}+4)}{2({n_0}+2)({n_0}+1)({n_0}+2)}$, which is equal to $0$.
I have done everything to my knowledge but am still stuck. Any tips?
 A: Your sum equals $0$.
The sum can be simplified using the idea of telescoping series.
$$\begin{align}&\sum_{i=1}^{n_1} \frac{(-1)^{i+1}2^{{n_1}-i}i({n_0}({n_1}-2+i)+{n_1}(i-1)+i^2-i)}{i!({n_1}-i)!({n_0}+i)({n_0}+i-1)({n_0}+{n_1})}
\\\\&=\frac{-2^{n_1}}{n_0+n_1}\sum_{i=1}^{n_1} \frac{(-\frac 12)^{i}({n_0}({n_1}-2+i)+{n_1}(i-1)+i^2-i)}{(i-1)!({n_1}-i)!({n_0}+i)({n_0}+i-1)}\tag1
\\\\&=\frac{-2^{n_1}}{n_0+n_1}\sum_{i=1}^{n_1}(f(i+1)-f(i))
\\\\&=\frac{-2^{n_1}}{n_0+n_1}(f(n_1+1)-f(1))
\\\\&=0\end{align}$$
where
$$f(i)=\frac{(-\frac 12)^{i-1}(i-1)(n_1+1-i)}{(i-1)!(n_1+1-i)!(n_0+i-1)}$$

Added : 

how does one find f(i)?

The denominator of the sum in $(1)$ is $(i-1)!({n_1}-i)!({n_0}+i)({n_0}+i-1)$.
To have this denominator in $f(i+1)-f(i)$, one can think that the denominator of $f(i)$ is $(i-2)!(n_1-i)!(n_0+i-1)$. This is because the denominator of $f(i+1)$ is $(i-1)!(n_1-i-1)!(n_0+i)$, so the denominator of $f(i+1)-f(i)$ can be $(i-1)!({n_1}-i)!({n_0}+i)({n_0}+i-1)$.
Having $(-\frac 12)^i$, one can think that $f(i)$ is of the form
$$\frac{(-\frac 12)^{i-1}g(i)}{(i-2)!(n_1-i)!(n_0+i-1)}$$
where $g(i)$ is a polynomial on $i$.
Finally, comparing the terms gives $g(i)=1$ fortunately, one has
$$f(i)=\frac{(-\frac 12)^{i-1}}{(i-2)!(n_1-i)!(n_0+i-1)}=\frac{(-\frac 12)^{i-1}(i-1)(n_1+1-i)}{(i-1)!(n_1+1-i)!(n_0+i-1)}$$
