summation - 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: $m≤n$

Answer

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
$$x_1+x_2+x_3+\cdots+x_{k+2}=n+2$$