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