Sum of binomial coefficients when lower suffices is same in the series: ${m \choose m}+{m+1 \choose m}+{m+2 \choose m}+...+{n \choose m}$ I want to find out sum of the following series:
$${m \choose m}+{m+1 \choose m}+{m+2 \choose m}+...+{n \choose m}$$
My try:
${m \choose m}+{m+1 \choose m}+{m+2 \choose m}+...+{n \choose m}$ = Coefficient of $x^m$ in the expansion of $(1+x)^m + (1+x)^{m+1} + ... + (1+x)^n$
 Or, Coefficient of $x^m$
$$\frac{(1+x)^{m}((1+x)^{n}-1)}{1+x-1}$$
$$=\frac{(1+x)^{m+n}-(1+x)^{m}}{x}$$
But, how to proceed further?
Note: $m≤n$
 A: The coefficient of $x^m$ in
$$\frac{(1+x)^{n+1}-(1+x)^{m}}{x}$$
is obviously equal to the coefficient of $x^{m+1}$ in 
$$(1+x)^{n+1}-(1+x)^{m}$$
which is nothing but
$${n+1\choose m+1},$$
as $n\geq m$ and the second term has degree $m<m+1$ and so gives no contribution.
A: Other way
$$\left( \begin{matrix}
   k  \\
   m  \\
\end{matrix} \right)=\left( \begin{matrix}
   k+1  \\
   m+1  \\
\end{matrix} \right)-\left( \begin{matrix}
   k  \\
   m+1  \\
\end{matrix} \right)
$$
we have
$$\sum\limits_{k=m}^{n}{\left( \begin{matrix}
   k  \\
   m  \\
\end{matrix} \right)}=\sum\limits_{k=m}^{n}\left[{\left( \begin{matrix}
   k+1  \\
   m+1  \\
\end{matrix} \right)-\left( \begin{matrix}
   k  \\
   m+1  \\
\end{matrix} \right)}\right]=\left( \begin{matrix}
   n+1  \\
   m+1  \\
\end{matrix} \right)
$$
 Also
Let $x_i\in \mathbb{N}$ and
$$x_1+x_2+x_3+\cdots+x_{k+2}=n+2$$
A: It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n
\end{align*}

We obtain for $0\leq m \leq n$
\begin{align*}
\sum_{k=m}^{n}\binom{k}{m}
&=\sum_{k=m}^n[x^m](1+x)^k\tag{1}\\
&=[x^m]\sum_{k=m}^n(1+x)^k\tag{2}\\
&=[x^m]\frac{(1+x)^{m}-(1+x)^{n+1}}{-x}\tag{3}\\
&=[x^{m+1}]\left((1+x)^{n+1}-(1+x)^{m}\right)\tag{4}\\
&=\binom{n+1}{m+1}\tag{5}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we apply the coefficient of operator.

*In (2) we use the linearity of the coefficient of operator.

*In (3) we use the finite geometric series formula.

*In (4) we do some simplifications and  we also  use  the  rule
\begin{align*}
[x^{p+q}]A(x)=[x^p]x^{-q}A(x)
\end{align*}

*In (5) we select the coefficient of $x^{m+1}$ and observe that $(1+x)^{m}$ has no contribution.
