Is there an explicit expression for $\sum_{j= k+1}^{2n}{j\choose k}{n\choose j-n}$? \begin{align*}\text{Let     } \ \ \    \ \   a_{n,k}:=\sum_{j= k+1}^{2n}{j\choose k}{n\choose j-n}\end{align*}
with  ${a\choose b}$ a binomial coefficient.
(i) Show that $a_{n,k}$ is even. 
[EDIT: I have just found a proof for (i). Hint: $a_{n,k}$ is the coefficient of $x^{2n-k}$ in the power series expansion of $\frac{(1+x)^n}{(1-x)^{k+1}}-(1+x)^n$.  ] 

2nd EDIT: here is my tentative proof with details (is it
  correct?):\begin{align*}a_{n,k}&=\sum_{j= k+1}^{2n}{j\choose
k}{n\choose j-n}\\ & =\sum_{j= k}^{2n}{j\choose k}{n\choose
2n-j}-{n\choose 2n-k}\\ &=\sum_{j= 0}^{2n-k}{j+k\choose k}{n\choose
2n-k-j}-{n\choose 2n-k}\end{align*} On the other hand, it is well
  known that:
$\sum_{g\ge0}{g+k\choose k}x^g=(1-x)^{-k-1}$ and
  $\sum_{g\ge0}{n\choose g}x^g=(1+x)^n$, so that:
  $$\frac{(1+x)^n}{(1-x)^{k+1}} -(1+x)^n = \sum_{h \ge 0} \sum_{g=
0}^{h}{g+k\choose k}{n\choose h-g}x^h -\sum_{h \ge 0}{n\choose h}x^h
$$  Then  $$[[x^{2n-k}]]\left(\frac{(1+x)^n}{(1-x)^{k+1}} -(1+x)^n
\right) = a_{n,k} $$ Then, reducing modulo $2$:
  $$[[x^{2n-k}]]\left((1-x)^{n-k-1} -(1+x)^n \right) \equiv a_{n,k}
\pmod 2.$$ There are two cases:
$\circ$ If $k+1\le n$  then $2n-k \ge n +1$, hence $2n-k > n $  and 
  $2n-k > n -k-1 \ge 0 $, hence:   \begin{align*} a_{n,k}&\equiv (-1)^k
{n-k-1\choose 2n-k} -{n\choose 2n-k} \pmod 2 \\ &\equiv 0-0=0\pmod
2.\end{align*}  $\circ$ If $0 > n-k-1$ , then:   \begin{align*}  a_{n,k}&\equiv  {k-n+2n-k\choose 2n-k} -{n\choose 2n-k} \pmod 2 \\
&\equiv {n\choose 2n-k} -{n\choose 2n-k}=0\pmod 2.\end{align*}

(ii) Give an explicit expression for $a_{n,k}$.
I know that \begin{align*}\sum_{j= k}^{n}{j\choose k}{n\choose j} =2^{n-k}{n\choose k}\end{align*} and I would expect some similar expression for $a_{n,k}$, but that does not seem to be straightforward.
Thank you for any help.
 A: Regarding your question for correctness: your proof looks fine, good work.

In this answer we derive another representation for $a_{n,k}$ and consider analogies with OPs  referred binomial identity
\begin{align*}
\sum_{j= k}^{n}{j\choose k}{n\choose j} =2^{n-k}{n\choose k}\tag{1}
\end{align*}
which indicate that a closed representation similarly to (1) is not to expect.

We show the following is valid for $n\geq 1,\ 0\leq k\leq 2n-1$
  \begin{align*}
\color{blue}{a_{n,k}}&=\sum_{j= k+1}^{2n}\binom{j}{k}\binom{n}{j-n}\tag{2}\\
&\color{blue}{=2^{n-k}\sum_{j=0}^k\binom{n}{j}\binom{n}{k-j}2^j-\binom{n}{k-n}}\tag{3}
\end{align*}
Note that $a_{n,k}$ has according to (3) also a representation with $2^{n-k}$ as factor similarly to (1), but then followed by a sum instead of a plain binomial coefficient. Subtraction of $\binom{n}{k-n}$ compensates for the index $j$ in (2) which starts with $k+1$ and not with $k$.

In the following we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{j=k}^{2n}\binom{j}{k}\binom{n}{j-n}}&=\sum_{j=0}^{2n-k}\binom{k+j}{k}\binom{n}{j+k-n}\tag{4}\\
&=\sum_{j=0}^{2n-k}\binom{2n-j}{k}\binom{n}{j}\tag{5}\\
&=\sum_{j=0}^\infty[z^k](1+z)^{2n-j}[u^j](1+u)^n\tag{6}\\
&=[z^k](1+z)^{2n}\sum_{j=0}^\infty(1+z)^{-j}[u^j](1+u)^n\tag{7}\\
&=[z^k](1+z)^{2n}\left(1+\frac{1}{1+z}\right)^n\tag{8}\\
&=[z^k](1+z)^n(2+z)^n\tag{9}\\
&=[z^k]\sum_{j=0}^n\binom{n}{j}z^j(2+z)^n\\
&=\sum_{j=0}^k\binom{n}{j}[z^{k-j}](2+z)^n\\
&=\sum_{j=0}^k\binom{n}{j}\binom{n}{k-j}2^{n-k+j}\\
&\color{blue}{=2^{n-k}\sum_{j=0}^k\binom{n}{j}\binom{n}{k-j}2^{j}}\tag{10}
\end{align*}
  Subtracting $\binom{n}{k-n}$ from (10)  to compensate for $j=k+1$  instead of $j=k$ in (2) and the claim follows.

Comment:


*

*In (4)  we shift  the index to start with $j=0$.

*In (5) we change the order of summation $j \rightarrow 2n-k-j$ and use $\binom{p}{q}=\binom{p}{p-q}$.

*In (6) we apply the coefficient of operator twice and set the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.

*In (7) we do a rearrangement to prepare for the next step.

*In (8) we apply the substitution rule of the coefficient of operator with $u=\frac{1}{1+z}$
\begin{align*}
A(z)=\sum_{j=0}^\infty a_j z^j=\sum_{j=0}^\infty z^j [u^j]A(u)
\end{align*}

*In (9) we do a simplification and select the coefficient of $z^k$ in the following steps.

The other binomial identity
We consider OPs referred binomial identity 
  \begin{align*}
\color{blue}{\sum_{j= k}^{n}{j\choose k}{n\choose j} =2^{n-k}{n\choose k}}\tag{11}
\end{align*} and compare it with $a_{n,k}$. 

This identity is usually shown by noting that 
$\binom{j}{k}\binom{n}{j}=\binom{n}{k}\binom{n-k}{n-j}$
from which the claim easily follows by factoring out $\binom{n}{k}$ from the sum. Here we prove it by proceeding similarly as we did above.

We obtain
  \begin{align*}
\color{blue}{\sum_{j=k}^n\binom{j}{k}\binom{n}{j}}&=\sum_{j=0}^{n-k}\binom{k+j}{k}\binom{n}{k+j}\\
&=\sum_{j=0}^{n-k}\binom{n-j}{k}\binom{n}{j}\\
&=\sum_{j=0}^\infty[z^k](1+z)^{n-j}[u^j](1+u)^n\\
&=[z^k](1+z)^n\sum_{j=0}^\infty(1+z)^{-j}[u^j](1+u)^n\\
&=[z^k]\color{blue}{(1+z)^n}\left(1+\frac{1}{1+z}\right)^n\tag{12}\\
&=[z^k](2+z)^n\\
&\color{blue}{=2^{n-k}\binom{n}{k}}
\end{align*}
Conclusion: In (12) we see the difference between the power series expansion for this binomial sum which has a term $(1+z)^{\color{blue}{n}}$ while the power series expansion for $a_{n,k}$ has a term $(1+z)^{\color{blue}{2n}}$. While $(1+z)^n$ cancels in the next step this is not the case in (9). This strongly indicates that a simpler representation of $a_{n,k}$ similarly as we found here is not feasible.

