How can I prove the following identity:
$$\sum_{0\le i\le n} (i+1)2^i {{2n-i}\choose{n}} =(2n+1){{2n}\choose{n}}$$
A few words as to why I need the proof.
I noticed that if we write numbers of the form $n2^{n-1}$ (so, 0, 1, 4, 12, 32 etc.) on the sides of Pascal triangle and continue to build Pascal triangle by the same rule as usually, then numbers $n{{2n}\choose{n}}$ (2, 12, 60, etc.) emerge on the diagonal.
Here is the picture showing ratios of the numbers in the new triangle to the corresponding numbers in the usual Pascal triangle
One can write down the generating function with two variables for the entries in our new triangle. Let $a_{i,j}$ be the entry in the i-th column and j-th row, then $$F(x,y) = \sum a_{i, j} x^i y^j = \frac{ \frac{x-x^2}{(1-2x)^2} + \frac{y-y^2}{(1-2y)^2} }{1-x-y}$$
From here I obtained the aforementioned identity comparing the coefficients of $x^n y^n$.
But I don't know how to prove it. The upper coefficient in the sum depends on $i$, so this formula has no algebraic meaning. Induction doesn't help too.
Any hints are welcome.