How to calculate $\sum_{m=0}^{i} (-1)^m\binom{2i}{i+m}=\frac{1}{2}\binom{2i}{i}$ 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.
 A: 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 $i=2$:
$$\sum_{m=0}(-1)^m\binom4{m+2}=\binom42-\binom43+\binom44=6-4+1=3=\binom32\;.$$
A: 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 $i=2k+1$
$$
\eqalign{
& Denote\ \ \ a_m =(-)^{m+1}\binom{2i}{i+m} \ \ \ S_{2i}=\sum_{m=0}^{2i}a_m \ \ \ S_{i}=\sum_{m=0}^{i}a_m \ \ \ then \cr
S_{2i} &= 2S_i -(-)^i\binom{2i}{i} = 0 \cr
& \rightarrow  S_i = {(-)^i\over 2}\binom{2i}{i}= -{1\over 2}\binom{2i}{i}\ \ \ because\ i\ odd \cr
& \rightarrow \sum_{m=0}^{i}(-)^{m+1}\binom{2i}{i+m} = -{1\over 2}\binom{2i}{i} \cr
& \rightarrow \sum_{m=0}^{i}(-)^{m}\binom{2i}{i+m} = {1\over 2}\binom{2i}{i} \cr
}
$$
A: $$\begin{align}
(1+x)^{2n}&=\sum_{r=0}^{2n}\binom{2n}rx^r\\
&=\sum_{s=0}^{n}\binom {2n}s x^s+\sum_{s=0}^{n}\binom{2n}{n+s}x^{n+s}-\binom {2n}n x^n
\qquad\quad\small\text{to remove duplicate }\binom {2n}n x^n\\
&=\sum_{s=0}^{n}\left[\binom{2n}{n-s}x^{n-s}+\binom{2n}{n+s}x^{n+s}\right]-\binom {2n}nx^n
\qquad\small\text{counting backward and forward from }n\\
&=x^n\biggr\lbrace\sum_{s=0}^{n}\left[\binom{2n}{n-s}x^{-s}+\binom{2n}{n+s}x^{s}\right]-\binom {2n}n\biggr\rbrace\\
\text{Put } x=-1:\\
0&=(-1)^n\biggr\lbrace\sum_{s=0}^n\left[\binom {2n}{n-s}(-1)^{-s}+\binom{2n}{n+s}(-1)^s\right]-\binom {2n}n\biggr\rbrace\\
&=(-1)^n\underbrace{\biggr\lbrace 2\sum_{s=1}^n \binom{2n}{n+s}(-1)^s-\binom {2n}n\biggr\rbrace}_{=0}
\qquad\qquad\small\text{as} \binom {2n}{n-s}=\binom {2n}{n+s}\text{ and } (-1)^{-s}=(-1)^s\\
\sum_{s=0}^n(-1)^s\binom {2n}{n+s}&=\frac 12\binom {2n}n\\
\Rightarrow\sum_{m=0}^i(-1)^m\binom {2i}{i+m}&=\frac 12\binom {2i}i\qquad\blacksquare\\
\end{align}$$

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.
A: 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 $[x^i]$
