Prove that $\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0. $ 
Let $n$ be a positive integer. Prove that
  $$
\sum_{k=0}^{2n} \binom {2n+k}{k} \binom{2n}{k} \frac{(-1)^k}{2^k} \frac{1}{k+1} = 0.
$$

I am trying to solve this by using induction on $n$. I have proven the sum to be zero in the case $n=1$. Assuming that the sum is zero for $n=m$ ($m$ is a positive integer), how do I prove that it implies that the sum is zero for $n=m+1$?
Can I get some hints?

Starting from
$$\sum_{k=0}^{2n} {2n+k\choose k} {2n\choose k}
\frac{(-1)^k}{2^k} \frac{1}{k+1}$$
we get
$$\frac{1}{2n} \sum_{k=0}^{2n} {2n+k\choose k+1} {2n\choose k}
\frac{(-1)^k}{2^k}
= \frac{1}{2n} \sum_{k=0}^{2n} {2n+k\choose 2n-1} {2n\choose k}
\frac{(-1)^k}{2^k}
\\ = \frac{1}{2n} \sum_{k=0}^{2n} {2n\choose k}
\frac{(-1)^k}{2^k} [z^{2n-1}] (1+z)^{2n+k}
\\ = \frac{1}{2n} [z^{2n-1}] (1+z)^{2n}
\sum_{k=0}^{2n} {2n\choose k} 
\frac{(-1)^k}{2^k} (1+z)^{k}
\\ = \frac{1}{2n} [z^{2n-1}] (1+z)^{2n}
\left(1-\frac{1}{2} (1+z)\right)^{2n}
\\ = \frac{1}{2^{2n+1}n} [z^{2n-1}] (1+z)^{2n} (1-z)^{2n}
= \frac{1}{2^{2n+1}n} [z^{2n-1}] (1-z^2)^{2n} = 0.$$
The  last  step is  zero  by  inspection  since  we are  extracting  a
coefficient on an odd power from a polynomial where all the powers are
even, and we have the claim.
