TITLE: Solving $\sum_{k=0}^n 2^k \binom{2n-k}{n}=4^n$ using generating functions
QUESTION [5 upvotes]: So I have to show this identity. $$\sum_{k=0}^n 2^k \binom{2n-k}{n}=4^n$$
I tried two techniques to do it. First one would be seeing that it's convolution of two sequences: $\langle 2^n \rangle _{n=0}^\infty$ and $\langle \binom{2n}{n} \rangle _{n=0}^\infty$
That gives me:
$$\lbrack \sum_{n=0}^\infty 2^n x^n \rbrack \times \lbrack \sum_{n=0}^\infty \binom{2n}{n} x^n \rbrack =$$
$$\frac{1}{1-2x} \  \frac{1}{\sqrt{1-4x}} =$$
$$\sum_{n=0}^\infty x^n \sum_{k=0}^n 2^k \binom{2n-k}{n-k}$$
So now we need to show that
$$\lbrack x^n \rbrack \frac{1}{1-2x} \frac{1}{\sqrt{1-4x}} = \frac{1}{1-4x}$$ But i'm unable to do it.
Second method would be snake oil method:
$$\sum_{n=0}^\infty x^n \sum_{k=0}^n 2^k \binom{2n-k}{n} =$$
$$\sum_{k=0}^\infty 2^k \sum_{n=k}^\infty x^n \binom{2n-k}{n} = \text{where l := n - k}$$
$$\sum_{k=0}^\infty 2^k x^k \sum_{l=0}^\infty x^l \binom{2l + k}{l+k} $$
But i'm stuck here too.
I mean i know that
$$\sum_{n=0}^\infty \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}}$$
And if i multiply it by $x^k$ i get
$$\sum_{n=0}^\infty \binom{2n+2k}{n+k} x^{n+k} = \frac{x^k}{\sqrt{1-4x}}$$
And it's almost what i need to calculate... but yeah, almost.
I will really appreciate some help on this.

REPLY [7 votes]: Note: [2017-11-07] With respect to @robjohn's comment: There's always more than one way to do it. Here are some more answers which might be interesting.

Proving $4^n=\sum_{k=0}^n2^k\binom{2n-k}{n}$
How to prove that $\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$.


Here is a variation based upon the coefficient of method. We use the symbol $[z^n]$ to denote the coefficient of $z^p$ in a series. This way we can write e.g.
\begin{align*}
  [z^p](1+z)^q=\binom{q}{p}
  \end{align*}

We obtain
  \begin{align*}
\sum_{k=0}^n\binom{2n-k}{n}2^k&=\sum_{k=0}^\infty[z^n](1+z)^{2n-k}2^k\tag{1}\\
&=[z^n](1+z)^{2n}\sum_{k=0}^\infty\left(\frac{2}{1+z}\right)^k\tag{2}\\
&=[z^n](1+z)^{2n}\frac{1}{1-\frac{2}{1+z}}\tag{3}\\
&=[z^n]\frac{(1+z)^{2n+1}}{z-1}\\
&=[z^n]\frac{1}{z}\sum_{j=0}^\infty z^{-j}(1+z)^{2n+1}\tag{4}\\
&=\sum_{j=0}^n[z^{n+1+j}](1+z)^{2n+1}\tag{5}\\
&=\sum_{j=0}^n\binom{2n+1}{n+1+j}\tag{6}\\
&=\sum_{j=0}^n\binom{2n+1}{j}\tag{7}\\
&=\frac{1}{2}2^{2n+1}\\
&=4^n
\end{align*}
  and the claim follows.

Comment:

In (1) we apply the coefficient of operator and extend the upper limit of the sum to $\infty$ without changing anything since we are adding zeros only.
In (2) we do some rearrangements and use the linearity of the coefficient of operator.
In (3) and (4) we use the geometric series expansion.
In (5) we use the rule $[z^{p+q}]A(z)=[z^p]z^{-q}A(z)$ and restrict the upper limit of the sum with $n$ since the exponent $n+1+j\leq2n+1$.
In (6) we select the coefficient of $z^{n-j}$.
In (7) we replace $j$ with $n-j$ and apply the binomial identity $\binom{p}{q}=\binom{p}{q-p}$.

Hint: The Cauchy product in your approach is slightly different, since
\begin{align*}
\left(\sum_{n=0}^\infty 2^nx^n\right)\left(\sum_{n=0}^\infty\binom{2n}{n}x^n\right)
=\sum_{n=0}^\infty\left(\sum_{k=0}^n2^k\binom{2n\color{blue}{-2k}}{n-k}\right)x^n
\end{align*}