combinatorics summation problem My problem is following:
$$\binom{n}{r} + \binom{n+1}{r+1} + \binom{n+2}{r+2} + \dots + \binom{n+M}{r+M}$$
how can we reduce it to a more short solution
Here $\dbinom{n}{r} = \dfrac{n!}{r! (n-r)!}$ and thus same as regular.Please help me in solving the above expression
 A: Hint what happens if you add :$\dbinom{n}{r-1}$ to your sum, can you prove that it will be simplified recursively using $$\dbinom{n}{k}+\dbinom{n}{k+1}=\dbinom{n+1}{k+1}$$
Answer $\dbinom{n+M+1}{r+M}-\dbinom{n}{r-1}$
A: $$\begin{align*}
\sum_{k=0}^m\binom{n+k}{r+k}&=\sum_{k=0}^m\binom{n+k}{n-r}\\
&=\sum_{\ell=r}^{m+r}\binom{n-r+\ell}{n-r}\\
&=\sum_{\ell=0}^{m+r}\binom{n-r+\ell}{n-r}-\sum_{\ell=0}^{r-1}\binom{n-r+\ell}{n-r}\\
&=\binom{n+m+1}{n-r+1}-\binom{n}{n-r+1}\\
\end{align*}$$
by a standard identity.
A: This question can  be treated using basic complex  variables, which is
an instructive exercise.
Suppose we seek to compute
$$\sum_{q=0}^M {n+q\choose r+q}$$
with $n\ge r.$
Introduce the integral representation
$${n+q\choose r+q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+q}}{z^{r+q+1}} \; dz.$$
This gives for the sum 
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r+1}} 
\sum_{q=0}^M \frac{(1+z)^q}{z^q}\; dz.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r+1}} 
\frac{(1+z)^{M+1}/z^{M+1}-1}{(1+z)/z-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r}} 
\frac{(1+z)^{M+1}/z^{M+1}-1}{(1+z)-z} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r}} 
\left((1+z)^{M+1}/z^{M+1}-1\right) \; dz.$$
This has two pieces, the first is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+M+1}}{z^{r+M+1}} \; dz
= {n+M+1\choose r+M}$$
and the second is
 $$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r}} \; dz
= {n\choose r-1}.$$
Combine these two to get
$${n+M+1\choose r+M} - {n\choose r-1}.$$
