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?
$$\sum_{n=0}^N\binom{2N-n}N2^n(n+1)=(1+2N)\binom{2N}N$$
 A: $$\begin{align}
\sum_{n=0}^N\binom {2N-n}N2^n(n+1)
&=\sum_{n=0}^N\binom {2N-n}N\sum_{j=0}^n \binom nj(n+1)
&&\scriptsize\text{using }\sum_{j=0}^n \binom nj=2^n\\
&=\sum_{n=0}^N\binom {2N-n}N\sum_{j=0}^n \binom {n+1}{j+1}(j+1)\\
&=\sum_{n=0}^N\binom {2N-n}N\sum_{j=1}^{n+1} \binom {n+1}{j}j\\
&=\sum_{j=1}^{N+1}j\sum_{n=0}^{j-1}\binom {2N-n}N\binom {n+1}j
&&\scriptsize (0\le n<j\le N+1)\\
&=\sum_{j=1}^{N+1}j\binom {2N+2}{N+1+j}
&&\scriptsize \text{using}\sum_n\binom {a-n}b\binom {c+n}d=\binom{a+c+1}{b+d+1}\\
&=\frac 12(N+1)\binom {2N+2}{N+1}
&&\scriptsize\text{using (*) }\\
&=\frac 12(N+1)\cdot \frac {2N+2}{N+1}\cdot \binom {2N+1}{N}\\
&=(N+1)\cdot \binom {2N+1}{N+1}\\
&=(N+1)\cdot\frac {2N+1}{N+1}\cdot \binom {2N}N\\
&=\color{red}{(1+2N)\binom {2N}N}\qquad \blacksquare
\end{align}$$

*See derivation below. Putting $n=N+1$ gives the result used above. 
$$\small\begin{align}
\sum_{r=1}^n\binom{2n}{n+r}r
&=\sum_{j=n+1}^{2n}\binom {2n}j(j-n)\\
&=\sum_{j=n+1}^{2n}\binom {2n}jj-n\sum_{j=n+1}^{2n}\binom {2n}j\\
&=n2^{2n-1}-n\cdot \frac 12\left(\left(\sum_{j=0}^{2n}\binom {2n}j\right)-\binom {2n}n\right)\\
&=n2^{2n-1}-\frac 12n\left(2^{2n}-\binom {2n}n\right)&&
\hspace{2.5cm}\\
&=\frac 12n\binom {2n}n\end{align}$$
Note that 
$$\begin{align}
\frac 12(n+1)\binom {2n}{n+1}
&=\frac 12 (n+1)\frac {(2n)!}{(n+1)!(n-1)!}\\
&=\frac 12 \cdot \frac {(2n)!}{n!(n-1)!}\cdot\color{grey}{ \frac nn}
\qquad\hspace{3cm}\\
&=\frac 12 n\cdot \frac {(2n)!}{n!n!}\\
&=\frac 12 n\binom {2n}n
\end{align}$$

Note also that 
$$\small\begin{align}
\sum_{n}\binom {a-n}b\binom {c+n}d
&=\sum_n\binom {a-n}{a-b-n}\binom {c+n}{c+n-d}\\
&=\sum_n(-1)^{a-b-n}\binom {-b-1}{a-b-n}(-1)^{c+n-d}\binom {-d-1}{c+n-d}
&&\text{(upper negation)}\\
&=(-1)^{a-b+c-d}\sum_n\binom {-b-1}{a-b-n}\binom {-d-1}{c-d+n}\\
&=(-1)^{a-b+c-d}\binom {-b-d-2}{a-b+c-d}
&&\text{(Vandermonde)}\\
&=(-1)^{a-b+c-d}(-1)^{a-b+c-d}\binom {a+c+2-1}{a-b+c-d}
&&\text{(upper negation)}\\
&=\binom {a+c+1}{a-b+c-d}\\
&=\binom {a+c+1}{b+d+1}\end{align}$$
A: 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.
