Sum of series ${n\choose 2a}{a\choose 0}+ {n\choose {2a+2}}{{a+1}\choose 1} + {n\choose {2a+4}}{{a+2}\choose 2} + \ldots$ I wanted to check the rationality of the cosine function for some rational multiples of $\pi$. And I found out that, $\cos(n \cdot\arccos x)$ generates a polynomial in $x$ whose co-efficients have the form:
$${n\choose 2a}{a\choose 0}+ {n\choose {2a+2}}{{a+1}\choose 1} + {n\choose {2a+4}}{{a+2}\choose 2} + \ldots$$
For $a = 0$ the answer is simple, but what for other cases? Please help.
 A: Suppose we seek to evaluate
$$\sum_{k=0}^{n} {n\choose 2a+2k} {a+k\choose k}$$
where $n\ge a\ge 0.$
Introduce
$${n\choose 2a+2k} = {n\choose n-2a-2k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a-2k+1}} 
(1+z)^{n} \; dz.$$
Observe  that this is  zero when  $2k\gt n-2a$  which is  the correct
value.  Therefore we may extend the range of $k$ to infinity to obtain
for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}} 
(1+z)^{n} 
\sum_{k\ge 0} {a+k\choose a} z^{2k}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}} 
(1+z)^{n} \frac{1}{(1-z^2)^{a+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}} 
(1+z)^{n-a-1} \frac{1}{(1-z)^{a+1}} \; dz.$$
Extracting the residue we get
$$\sum_{q=0}^{n-2a} {n-a-1\choose q} 
{n-2a-q+a\choose a} =
\sum_{q=0}^{n-2a} {n-a-1\choose q} 
{n-a-q\choose a}
\\ = \sum_{q=0}^{n-2a} {n-a-1\choose q} 
{n-a-q\choose n-2a-q}.$$
This time introduce
$${n-a-q\choose n-2a-q} 
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a-q+1}} 
(1+z)^{n-a-q} \; dz$$
which is zero when $q\gt n-2a$ so that we may extend $q$ to $n-a-1$ to
get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}} 
(1+z)^{n-a} 
\sum_{q=0}^{n-a-1} {n-a-1\choose q} \frac{z^q}{(1+z)^q}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}} 
(1+z)^{n-a} 
\left(1+\frac{z}{1+z}\right)^{n-a-1}
\; dz 
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1+z}{z^{n-2a+1}} 
\left(1+2z\right)^{n-a-1}
\; dz.$$
This yields the closed form
$$2^{n-2a} {n-a-1\choose n-2a} +  
2^{n-2a-1} {n-a-1\choose n-2a-1}
\\ = 2^{n-2a} \frac{a}{n-2a} {n-a-1\choose n-2a-1} +  
2^{n-2a-1} {n-a-1\choose n-2a-1}
\\ = \left(2 \frac{a}{n-2a} + 1\right)
2^{n-2a-1} {n-a-1\choose n-2a-1}
= \frac{n}{n-2a} 2^{n-2a-1} {n-a-1\choose n-2a-1}.$$
This holds when $2a\lt n.$  When $2a=n$ the initial closed form yields
$1$ because the second binomial coefficient vanishes.
A: $\text{The Chebyshev polynomial of the first kind is given by}$
$$\boxed{\color{red}{T_n(x) = \cos(n\arccos(x)) = \dfrac{n}2 \sum_{k \leq n/2}(-1)^k \dfrac{(n-k-1)!}{k!(n-2k)!}(2x)^{n-2k}}}$$
$\text{You can now compare coefficients and obtain the answer.}$
