Sum of binomial coefficients when lower suffices is same in the series: ${m \choose m}+{m+1 \choose m}+{m+2 \choose m}+...+{n \choose m}$

Question

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 1

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

Answer 2

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.

Answer 3

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.