Problem: Show that $P^{'}_{2n+1}(0)= \frac{(-1)^n (2n+1){^{2n}}C_n}{2^n}$.
My attempt: Given: $P_{n}(x)=\frac{1}{2^n}\sum_{k=0}^{\frac{1}{2}n}(-1)^{k} {{^n}}C_k{^{2n-2k}C_n}x^{n-2k}$
Let $n = 2n+1$. We have $P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{k=0}^{{n}+\frac{1}{2}}(-1)^{k} {{^{2n+1}}}C_k{^{2(2n+1)-2k}C_{2n+1}}x^{2n+1-2k}$
Now, let $k=n$.
$P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2(2n+1)-2n}C_{2n+1}}x^{2n+1-2n}$
$P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2n+2}C_{2n+1}}x^{1}$
$P_{2n+1}^{'}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2n+2}C_{2n+1}}$
Upon simplifying, we have
$P^{'}_{2n+1}(0)= \frac{(-1)^n (2n+1){^{2n}}C_n}{2^{2n}}$
This result is different from the one given. I have missed out something, please help!