How to prove this $\sum_{k=0}^{n}(-1)^k2^k\binom{2n-1}{k}\binom{2n-k}{n-k}=0$ prove that
$$\sum_{k=0}^{n}(-1)^k2^k\binom{2n-1}{k}\binom{2n-k}{n-k}=0$$
This problem is from  book, But I Think this book have some questions
I Think (2) $r$ replace $n$? is my true?

and How to show by $(2)$  Thank you 
 A: Let's try to simplify things here. We get
$$\sum_{k=0}^{n}(-1)^k2^k\binom{2n-1}{k}\binom{2n-k}{n-k}=\frac{(2n-1)!}{n!n!}\sum_{k=0}^{n}(-1)^k2^k\binom{ n }{k} (2n-k).$$
The first factor is useless for us. We continue
$$\sum_{k=0}^{n}(-1)^k2^k\binom{ n }{k} (2n-k)=2n\sum_{k=0}^{n}(-1)^k2^k\binom{ n }{k} -\sum_{k=0}^{n}k(-1)^k2^k\binom{ n }{k}$$
$$=2n (1-2)^n-\sum_{k=0}^{n}k(-1)^k2^k\binom{ n }{k}.$$
Now $$\sum_{k=0}^{n}k(-1)^k2^k\binom{ n }{k}=2n\sum_{k=1}^{n}(-1)^k2^{k-1}\frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}$$
$$=2n \sum_{l=0}^{n-1}(-1)^{l+1}2^{l}\binom{n-1}{l}=-2n \sum_{l=0}^{n-1}(-1)^{l}2^{l}\binom{n-1}{l}=-2n(1-2)^{n-1}=2n(1-2)^{n}.$$
Now we look at the total sum
$$2n (1-2)^n-2n(1-2)^{n}=0.$$
A: For $n\gt0$,
$$
\begin{align}
\sum_{k=0}^n(-1)^k2^k\binom{2n-1}{k}\binom{2n-k}{n-k}
&=\sum_{k=0}^n(-2)^k\binom{2n}{n}\color{#C00000}{\binom{n}{k}\frac{2n-k}{2n}}\\
&=\binom{2n}{n}\sum_{k=0}^n(-2)^k\left(\color{#C00000}{\binom{n}{k}-\frac12\binom{n-1}{k-1}}\right)\\
&=\binom{2n}{n}\left(\sum_{k=0}^n(-2)^k\binom{n}{k}+\sum_{k=0}^n(-2)^{k-1}\binom{n-1}{k-1}\right)\\
&=\binom{2n}{n}\left((1-2)^n+(1-2)^{n-1}\right)\\[6pt]
&=0
\end{align}
$$
