How can I prove this combinatorial identity $\sum_{j=0}^n j\binom{2n}{n+j}\binom{m+j-1}{2m-1}=m\cdot4^{n-m}\cdot\binom{n}{m}$? Let $n,m$ be non-negative integers.
How can one prove the following identity?

$$\sum_{j=0}^n j\binom{2n}{n+j}\binom{m+j-1}{2m-1}=m\cdot4^{n-m}\cdot\binom{n}{m}$$

 A: Let :
$$f_n:=\sum_{j=0}^n j\binom{2n}{n+j}\binom{m+j-1}{2m-1}$$
$$g_n:=m\cdot4^{n-m}\cdot\binom{n}{m}$$
I will do the "snake-oil method" (it is hardcore I must admit, have fun...) as seen in the following (wonderful) reference :
http://www.math.upenn.edu/~wilf/DownldGF.html
The idea, instead of showing $f_n=g_n$ for all $n$, take the associated formal power series $F$ and $G$ and show $F=G$, the snake oil method applies because $f_n$ is defined with a sum (I will invert the two sums and I will get the solution). 
Then set :
$$ F(X):=\sum_{n=0}^{+\infty}f_nX^n\text{ and }G(X):=\sum_{n=0}^{+\infty}g_nX^n$$
You have  :
$$G(X)=\sum_{n=0}^{+\infty}m\cdot4^{n-m}\cdot\binom{n}{m}X^n=\sum_{n=m}^{+\infty}m\cdot4^{n-m}\cdot\binom{n}{m}X^{n} $$
Then :
$$G(X)=\sum_{n=0}^{+\infty}m\cdot4^{n}\cdot\binom{n+m}{m}X^{n+m}=mX^m\sum_{n=0}^{+\infty}\binom{n+m}{m}(4X)^n=m\frac{X^m}{(1-4X)^{m+1}} $$
On the other hand :
$$F(X)=\sum_{n=0}^{+\infty}\sum_{j=0}^n j\binom{2n}{n+j}\binom{m+j-1}{2m-1}X^n=\sum_{j=0}^{+\infty}\sum_{n=j}^{+\infty} j\binom{2n}{n+j}\binom{m+j-1}{2m-1}X^n $$
$$F(X)=\sum_{j=0}^{+\infty}j\binom{m+j-1}{2m-1}\sum_{n=j}^{+\infty} \binom{2n}{n+j}X^n=\sum_{j=0}^{+\infty}j\binom{m+j-1}{2m-1}X^j\sum_{n=0}^{+\infty} \binom{2n+2j}{n+2j}X^n $$
And use the identity page 54 of the reference I have given to get :
$$\sum_{n=0}^{+\infty} \binom{2n+2j}{n}X^n=\frac{1}{\sqrt{1-4X}}(\frac{1-\sqrt{1-4X}}{2X})^{2j}$$
$$F(X)=\sum_{j=0}^{+\infty}j\binom{m+j-1}{2m-1}X^j\frac{1}{\sqrt{1-4X}}(\frac{1-\sqrt{1-4X}}{2X})^{2j}$$
Set $Y:=\sqrt{1-4X}$ then :
$$F(X)=\sum_{j=0}^{+\infty}j\binom{m+j-1}{2m-1}\frac{1}{(4X)^j}\frac{1}{Y}(1-Y)^{2j}$$
But $4X=1-Y^2$ so :
$$F(X)=\sum_{j=0}^{+\infty}j\binom{m+j-1}{2m-1}\frac{1}{(1-Y^2)^j}\frac{1}{Y}(1-Y)^{2j}$$
And then it reduces :
$$\frac{(1-Y)^{2j}}{(1-Y^2)^j} =\frac{(1-Y)^j}{(1+Y)^j}$$
$$F(X)=\frac{1}{Y}\sum_{j=0}^{+\infty}j\binom{m+j-1}{2m-1}Z^j\text{ where } Z:=\frac{1-Y}{1+Y}$$
This looks like a usual formal power series now :
$$F(X)=\frac{1}{Y}\sum_{j=m}^{+\infty}j\binom{m+j-1}{2m-1}Z^j=\frac{Z^m}{Y}\sum_{j=0}^{+\infty}(j+m)\binom{2m+j-1}{2m-1}Z^j$$
$$F(X)=\frac{Z^m}{Y}\sum_{j=0}^{+\infty}(j+m)\binom{2m+j-1}{j}Z^j$$
Now 
$$\sum_j \binom{2m+j-1}{j}Z^j=\frac{1}{(1-Z)^{2m}}$$
And 
$$\sum_j j\binom{2m+j-1}{j}Z^j=ZD(\frac{1}{(1-Z)^{2m}})=\frac{2mZ}{(1-Z)^{2m+1}} $$
Hence :
$$F(X)=\frac{Z^m}{Y}(\frac{m}{(1-Z)^{2m}}+\frac{2mZ}{(1-Z)^{2m+1}})=\frac{1}{Y}(\frac{Z}{(1-Z)^2})^mm\frac{1+Z}{1-Z} $$
By an easy calculation :
$$\frac{Z}{(1-Z)^2}=\frac{X}{Y^2}\text{ and }  \frac{1+Z}{1-Z}=\frac{1}{Y}$$
$$F(X)=\frac{X^mm}{(Y^2)^{m+1}}=G(X)$$
Formal power series are equal, hence the coefficients must be. I don't know if there exists a "better" answer to this.
