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

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
$${n+M+1\choose r+M} - {n\choose r-1}.$$
