<s>[INST] Prove that $\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$ I used WolframAlpha to calculate a sum but it didn't show me the way :( Anybody has a hint or a solution for proving this sum?
$$\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$$
 [/INST] Starting from
$$\sum_{n=0}^N {2N-n\choose N} 2^n (n+1)$$
we write
$$\sum_{n=0}^N {2N-n\choose N-n} 2^n (n+1)
= \sum_{n=0}^N 2^n (n+1) [z^{N-n}] (1+z)^{2N-n}
\\ = [z^N] \sum_{n=0}^N 2^n (n+1) z^n (1+z)^{2N-n}.$$
We may extend $n$ to infinity beyond $N$ because the sum term does not
contribute to the coefficient extractor in that case, getting
$$[z^N] (1+z)^{2N} \sum_{n\ge 0} 2^n (n+1) z^n (1+z)^{-n}
= [z^N] (1+z)^{2N} \frac{1}{(1-2z/(1+z))^2}
\\ = [z^N] (1+z)^{2N+2} \frac{1}{(1-z)^2}.$$
Extracting the coefficient we find
$$\sum_{q=0}^N {2N+2\choose q} (N+1-q).$$
The first piece here is
$$(N+1) \sum_{q=0}^N {2N+2\choose q}
= (N+1) \frac{1}{2} \left(2^{2N+2} - {2N+2\choose N+1}\right).$$
The second piece is
$$\sum_{q=1}^N {2N+2\choose q} q
= (2N+2) \sum_{q=1}^N {2N+1\choose q-1}
\\ = (2N+2) \sum_{q=0}^{N-1} {2N+1\choose q}
= (2N+2) \frac{1}{2} \left(2^{2N+1}
- {2N+1\choose N} - {2N+1\choose N+1} \right).$$
Joining the two pieces the powers of two cancel and we are left with
$$(2N+2) {2N+1\choose N} - \frac{1}{2} (N+1) {2N+2\choose N+1}
\\ = (2N+2) \frac{2N+1}{N+1} {2N\choose N}
- \frac{1}{2} (N+1) \frac{2N+2}{N+1} {2N+1\choose N}
\\ = 2 (2N+1) {2N\choose N}
- (N+1) \frac{2N+1}{N+1} {2N\choose N}
\\ = (2N+1) {2N\choose N}$$
as claimed.
</s>