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: $m≤n$

You made a mistake, result you should obtain is $(1+x)^m + (1+x)^{m+1} + ... + (1+x)^n=\frac{(1+x)^{n+1}-(1+x)^m}{x}$. Then, you can conclude that coefficient of $x^m$ is ${n+1 \choose m+1}$ — alans, Jul 30 '16 at 12:33
@alans I used the sum of a GP formula. Can you tell me exactly where am I going wrong? — Shuvam Shah, Jul 30 '16 at 12:40
I hope that the following will make it clear: $$(1+x)^m + (1+x)^{m+1} + ... + (1+x)^n=(1+x)^m(1+(1+x)+\cdots+(1+x)^{n-m})=(1+x)^m\frac{(1+x)^{n-m+1}-1}{1+x-1}=(1+x)^m\frac{(1+x)^{n-m+1}-1}{x}=\frac{(1+x)^{n+1}-(1+x)^m}{x}$$ — alans, Jul 30 '16 at 12:43
See also Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ and other posts linked there. — Martin Sleziak, Jan 22 '17 at 14:58

b00n heT · Answer 1 · 2016-07-30T12:34:12.403

1

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.

edited Jul 30 '16 at 12:34

answered Jul 30 '16 at 12:25

b00n heT

16,360
1
36
46

score 1 · Accepted Answer · answered Jul 30 '16 at 12:26

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

I was so close... Well thanks a lot... – Shuvam Shah Jul 30 '16 at 12:30 — Shuvam Shah, Jul 30 '16 at 12:30

Markus Scheuer · Answer 3 · 2016-07-30T18:13:52.333

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.

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

3 Answers3