<s>[INST] Closed form of a sum of binomial coefficients? I have the following function:

$T_n(d)=\sum\limits_{k=\frac{n-d}{2}}^{\lceil \frac{n}{2} \rceil}{k\choose \frac{n-d}{2}}$
  ${n \choose 2k}$, where $n,d\in \mathbb{N}^0$, and $n,d$ have the same parity.

Looking at the sequences for various $d$, it seems that the formula is a polynomial of degree $d$ in $n$. This is speculation, but for example, where defined, it would appear that:
$T_{n}(1)=n$, $T_{n}(2)=\frac{1}{2}n^2$, $T_{n}(3)=\frac{1}{6}n^3-\frac{1}{6}n$, and $T_{n}(4)=\frac{1}{24}n^4-\frac{1}{6}n^2$
In fact, after further numerical testing, it seems that 
$T_n(d)=\frac{n}{d!}\prod\limits_{j=1}^{d-1}(n-(2j-d))$
So my question is, is there a way to confirm the above results and show whether or not $T_n(d)$ is a polynomial of degree $d$ in $n$? If so, is there an intuitive reason why it is a polynomial with evenly spaced integer roots? The result seems rather elegant, if it's true.
EDIT: Simplifying the above expression, we have
$$T_n(d)=\frac{n}{d!}\prod\limits_{j=1}^{d-1}(n-(2j-d))=\frac{n}{d!}\frac{(n-1+(d-1))!!}{(n-1-(d-1))!!}=\frac{2^d n}{n+d}\cdot {\frac{n+d}{2} \choose d}$$
The third and fourth expressions in particular, seem like they would be very useful for a combinatorial proof. $n!!$ is the standard double factorial.
 [/INST] Another  useful  apparently   classic  technique  uses  basic  complex
variables.

Suppose we seek to evaluate
$$\sum_{k\ge m} {k\choose m} {2m+d\choose 2k}
= \sum_{k\ge m} {k\choose m} {2m+d\choose 2m+d-2k}$$
where  the second  binomial coefficient  serves to  delimit  the upper
bound of the range of $k$ which is finite.
Start from
$${2m+d\choose 2m+d-2k}
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{2m+d-2k+1}} (1+z)^{2m+d} \; dz.$$

This gives the following integral for the sum
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\sum_{k\ge m} {k\choose m}
\frac{1}{z^{2m+d-2k+1}} (1+z)^{2m+d} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon}  \frac{(1+z)^{2m+d}}{z^{2m+d+1}}
\sum_{k\ge m} {k\choose m} z^{2k} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon}  \frac{(1+z)^{2m+d}}{z^{2m+d+1}}
\sum_{k\ge 0} {k+m\choose m} z^{2k+2m} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon}  \frac{(1+z)^{2m+d}}{z^{d+1}}
\sum_{k\ge 0} {k+m\choose m} z^{2k} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon}  \frac{(1+z)^{2m+d}}{z^{d+1}}
\frac{1}{(1-z^2)^{m+1}} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon}  \frac{(1+z)^{m+d-1}}{z^{d+1}}
\frac{1}{(1-z)^{m+1}} \; dz.$$
Extracting coefficients we obtain
$$\sum_{q=0}^d {m+d-1\choose q} {d-q+m\choose m}.$$
Use another integral as in
$${d-q+m\choose m}
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{d-q+m} \; dz.$$
to obtain
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\sum_{q=0}^d {m+d-1\choose q}
\frac{1}{z^{m+1}} (1+z)^{d-q+m} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^{d+m}}{z^{m+1}}
\sum_{q=0}^d {m+d-1\choose q} (1+z)^{-q}\; dz.$$
The defining  integral is zero for $m+d-1 \ge q>d$  so we may extend  the sum to
$m+d-1$, getting
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^{d+m}}{z^{m+1}}
\sum_{q=0}^{m+d-1} {m+d-1\choose q} (1+z)^{-q}\; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^{d+m}}{z^{m+1}}
\left(1+\frac{1}{1+z}\right)^{m+d-1} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^{d+m}}{z^{m+1}}
\left(\frac{2+z}{1+z}\right)^{m+d-1} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1+z}{z^{m+1}}
(2+z)^{m+d-1} \; dz.$$
We may now extract coefficients to obtain
$$[z^m] (1+z) (2+z)^{m+d-1}
= [z^m] (2+z)^{m+d-1}
+ [z^{m-1} ] (2+z)^{m+d-1}
\\ = 2^{d-1} {m+d-1\choose m} 
+ 2^d {m+d-1\choose m-1}.$$
This needs some re-writing to make it match the other answer.
We get
$$2^{d-1} {m+d-1\choose d-1} 
+ 2^d {m+d-1\choose d}
\\ = 2^{d-1} \frac{d}{m+d} {m+d\choose d} 
+ 2^d \frac{m}{m+d} {m+d\choose d}
\\ = 2^d {m+d\choose d}
\left(\frac{1}{2} \frac{d}{m+d} + \frac{m}{m+d} \right)
\\ = 2^d {m+d\choose d} \frac{2m+d}{2m+2d}.$$

Interesting  exercise. There  is  another  one like  it  at this MSE link.
Observe that  these computations were  done without intervention  by a
computer except to check arithmetic.

A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
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