How to calculate $\sum_{m=0}^{i} (-1)^m\binom{2i}{i+m}=\frac{1}{2}\binom{2i}{i}$

Question

how can we calculate this?$$ \sum_{m=0}^{i} (-1)^m\binom{2i}{i+m}=\frac{1}{2}\binom{2i}{i} $$ It is alternating and contains the Binomial coefficients which are given in terms of factorials as, $$ \binom{2i}{i+m}= \frac{(2i)!}{i!(i-m)!} $$

Note it may be helpful to know that $$ \sum_{m=0}^{i} (-1)^m \binom{i}{m}=0 $$ as can be seen from just setting $x=1,y=-1$ in the formula $$(x+y)^i=\sum_{m=0}^i \binom{i}{m} x^{i-m}y^m$$ Thank you for your time.

Answer 1

Note that $\dbinom{2i}{i+m}=\dbinom{2i}{i-m}$ and $(-1)^i=(-1)^{-i}$, so

$$\begin{align*} 0&=\sum_{m=0}^{2i}(-1)^m\binom{2i}m\\ &=\sum_{m=0}^{i-1}(-1)^m\binom{2i}m+(-1)^i\binom{2i}i+\sum_{m=i+1}^{2i}(-1)^m\binom{2i}m\\ &=\sum_{m=0}^{i-1}(-1)^m\binom{2i}m+\binom{2i}i+(-1)^{m+i}\sum_{m=1}^i(-1)^m\binom{2i}{m+i}\\ &=\sum_{m=1}^i(-1)^{m-i}\binom{2i}{m-i}+(-1)^i\binom{2i}i+\sum_{m=1}^i(-1)^{m+i}\binom{2i}{m+i}\\ &=(-1)^i\sum_{m=1}^i(-1)^m\binom{2i}{m-i}+(-1)^i\binom{2i}i+(-1)^i\sum_{m=1}^i(-1)^m\binom{2i}{m+i}\\ &=(-1)^i\left(\binom{2i}i+2\sum_{m=1}^i(-1)^m\binom{2i}{m+i}\right)\;, \end{align*}$$

so

$$\binom{2i}i+2\sum_{m=1}^i(-1)^m\binom{2i}{m+i}=0\;,$$

and

$$\sum_{m=1}^i(-1)^m\binom{2i}{m+i}=-\frac12\binom{2i}i\;.$$

Thus,

$$\sum_{m=0}^i(-1)^m\binom{2i}{m+i}=\binom{2i}i-\frac12\binom{2i}i=\frac12\binom{2i}i=\frac12\cdot\frac{2i}i\binom{2i-1}{i-1}=\binom{2i-1}i\;.$$

As a sanity check let’s try $i=2$:

$$\sum_{m=0}(-1)^m\binom4{m+2}=\binom42-\binom43+\binom44=6-4+1=3=\binom32\;.$$

Answer 2

Formulating same ideas as in Brian's solution in a different way.

Case 1, let $i=2k$ $$ \eqalign{ & Denote\ \ \ a_m =(-)^m\binom{2i}{i+m} \ \ \ S_{2i}=\sum_{m=0}^{2i}a_m \ \ \ S_{i}=\sum_{m=0}^{i}a_m \ \ \ then \cr S_{2i} &= 2S_i -(-)^i\binom{2i}{i} = 0 \cr & \rightarrow S_i = {(-)^i\over 2}\binom{2i}{i}={1\over 2}\binom{2i}{i}\ \ \ because\ i\ even \cr & \rightarrow \sum_{m=0}^{i}(-)^m\binom{2i}{i+m} = {1\over 2}\binom{2i}{i} \cr } $$

Case 2, let $i=2k+1$ $$ \eqalign{ & Denote\ \ \ a_m =(-)^{m+1}\binom{2i}{i+m} \ \ \ S_{2i}=\sum_{m=0}^{2i}a_m \ \ \ S_{i}=\sum_{m=0}^{i}a_m \ \ \ then \cr S_{2i} &= 2S_i -(-)^i\binom{2i}{i} = 0 \cr & \rightarrow S_i = {(-)^i\over 2}\binom{2i}{i}= -{1\over 2}\binom{2i}{i}\ \ \ because\ i\ odd \cr & \rightarrow \sum_{m=0}^{i}(-)^{m+1}\binom{2i}{i+m} = -{1\over 2}\binom{2i}{i} \cr & \rightarrow \sum_{m=0}^{i}(-)^{m}\binom{2i}{i+m} = {1\over 2}\binom{2i}{i} \cr } $$

Answer 3

$$\begin{align} (1+x)^{2n}&=\sum_{r=0}^{2n}\binom{2n}rx^r\\ &=\sum_{s=0}^{n}\binom {2n}s x^s+\sum_{s=0}^{n}\binom{2n}{n+s}x^{n+s}-\binom {2n}n x^n \qquad\quad\small\text{to remove duplicate }\binom {2n}n x^n\\ &=\sum_{s=0}^{n}\left[\binom{2n}{n-s}x^{n-s}+\binom{2n}{n+s}x^{n+s}\right]-\binom {2n}nx^n \qquad\small\text{counting backward and forward from }n\\ &=x^n\biggr\lbrace\sum_{s=0}^{n}\left[\binom{2n}{n-s}x^{-s}+\binom{2n}{n+s}x^{s}\right]-\binom {2n}n\biggr\rbrace\\ \text{Put } x=-1:\\ 0&=(-1)^n\biggr\lbrace\sum_{s=0}^n\left[\binom {2n}{n-s}(-1)^{-s}+\binom{2n}{n+s}(-1)^s\right]-\binom {2n}n\biggr\rbrace\\ &=(-1)^n\underbrace{\biggr\lbrace 2\sum_{s=1}^n \binom{2n}{n+s}(-1)^s-\binom {2n}n\biggr\rbrace}_{=0} \qquad\qquad\small\text{as} \binom {2n}{n-s}=\binom {2n}{n+s}\text{ and } (-1)^{-s}=(-1)^s\\ \sum_{s=0}^n(-1)^s\binom {2n}{n+s}&=\frac 12\binom {2n}n\\ \Rightarrow\sum_{m=0}^i(-1)^m\binom {2i}{i+m}&=\frac 12\binom {2i}i\qquad\blacksquare\\ \end{align}$$

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NB - The solution above uses $n$ as the constant and $r, s$ as indices of summation. This avoids confusion which may arise subconsciously from using $i$ as a constant, as it is usually used as a summation index.

Answer 4

Here is another variation of the theme. We use the coefficient of operator $[x^i]$ to denote the coefficient of $x^{i}$ in a polynomial or series. We can write this way e.g. \begin{align*} \binom{i}{m}=[x^m](1+x)^i \end{align*}

We obtain \begin{align*} \sum_{m=0}^i&(-1)^m\binom{2i}{i-m}\tag{1}\\ &=\sum_{m=0}^{i}(-1)^m[x^{i-m}](1+x)^{2i}\tag{2}\\ &=[x^i](1+x)^{2i}\sum_{m=0}^{i}(-x)^m\tag{3}\\ &=[x^i](1+x)^{2i}\frac{1-(-x)^{i+1}}{1+x}\tag{4}\\ &=[x^i](1+x)^{2i-1}\left(1-(-x)^{i+1}\right)\tag{5}\\ &=[x^i](1+x)^{2i-1}\\ &=\binom{2i-1}{i}\\ &=\frac{1}{2}\binom{2i}{i} \end{align*}

Comment:

• In (1) we use $\binom{n}{k}=\binom{n}{n-k}$

• In (2) we use the coefficient of operator

• In (3) we rearrange the sum and use the rule $[x^{n+m}]A(x)=[x^n]x^{-m}A(x)$

• In (4) we use the finite geometric sum formula

• In (5) we observe that multiplication with $(-x)^{i+1}$ does nothing contribute to $[x^i]$