Combinatorial Summations!

Let $\displaystyle A_n = \sum_{k} \dbinom n{2k} \dbinom km$ , then

$\begin{aligned} \sum_{n=0}^\infty A_n x^n &=& \sum_{n=0}^\infty \left( \sum_k \dbinom n{2k} \dbinom km \right) x^n \\ &=& \sum_{k=0}^\infty \left( \sum_n \dbinom n{2k} x^n \right) \\ &=& \sum_{k=0}^\infty \dbinom km \dfrac{x^{2k}}{(1-x)^{2k+1}} \\ &=& \dfrac1{1-x} \sum_{k=0}^\infty \dbinom km \left( \dfrac{x^2}{(1-x)^2} \right)^k \\ &=& \dfrac1{1-x} \left [ \left( \dfrac{x^2}{(1-x)^2} \right)^m \div \left( 1 - \dfrac{x^2}{(1-x)^2} \right)^{m+1} \right ] \\ &=& (1-x) \dfrac{x^{2m}}{(1-2x)^{m+1}} \\ &=& (1-x) \left( \dfrac x2\right)^m \dfrac{(2x)^m}{(1-2x)^{m+1}} \\ &=& (1-x) \left( \dfrac x2\right)^m \sum_{r=m}^\infty \dbinom rm (2x)^r \end{aligned}$

By comparing the coefficients of $x^n$ we get that

$\begin{aligned} A_n &=& \dfrac1{2^m} \left [ \dbinom{n-m}m 2^{n-m} - \dbinom{n-m-1}m 2^{n-m-1}\right ] \\ &=& 2^{n-2m-1} \left [2\dbinom{n-m}m -\dfrac{n-2m}{n-m} \dbinom{n-m}m \right ] \\ &=& 2^{n-2m-1} \dfrac n{n-m} \dbinom{n-m}m \end{aligned}$

In this case, $n=40$ and $m=15$ . The result follows.