Proving $4^n=\sum_{k=0}^n2^k\binom{2n-k}{n}$

Question

$4^n = \sum\limits_{k=0}^{n}2^k\cdot{{2n - k} \choose n}$

I tried formal power series, but failed.

Answer 1

A solution through Complex Analysis and the residue theorem.

$$\mathcal{S}(n)=\sum_{k=0}^{n} 2^k\binom{2n-k}{n} = \sum_{k=0}^{n}2^{n-k}\binom{n+k}{n} $$ is the coefficient of $x^n$ in the product between $\sum_{k\geq 0}2^k x^k=\frac{1}{1-2x}$ (geometric series) and $\sum_{k\geq 0}\binom{n+k}{n}x^k = \frac{1}{(1-x)^{n+1}}$ (stars and bars). In particular

$$ \mathcal{S}(n)=\text{Res}\left(\frac{1}{(1-2x)\left[x(1-x)\right]^{n+1}},x=0\right) $$ but due to the symmetry of the meromorphic function $\frac{1}{(1-2x)\left[x(1-x)\right]^{n+1}}$ the residue at $0$ and the residue at $1$ are the same number. The only other pole is at $x=\frac{1}{2}$ and the sum of the residues is zero, hence $\mathcal{S}(n)=4^n$ can be proved from the straightforward $$ \text{Res}\left(\frac{1}{(1-2x)\left[x(1-x)\right]^{n+1}},x=\frac{1}{2}\right)=-2\cdot 4^n. $$

Answer 2

$$ \begin{align} \sum_{k=0}^n2^k\binom{2n-k}{n} &=\sum_{k=0}^n\sum_{j=0}^k\binom{k}{j}\binom{2n-k}{n}\tag1\\ &=\sum_{j=0}^n\sum_{k=j}^n\binom{k}{j}\binom{2n-k}{n}\tag2\\ &=\sum_{j=0}^n\binom{2n+1}{n+j+1}\tag3\\ &=\sum_{j=0}^n\binom{2n+1}{n-j}\tag4\\ &=\frac12\sum_{j=0}^{2n+1}\binom{2n+1}{j}\tag5\\ &=\frac12\cdot2^{2n+1}\tag6\\[12pt] &=4^n\tag7 \end{align} $$ Explanation:
$(1)$: use the binomial theorem to expand $(1+1)^n$
$(2)$: change order of summation
$(3)$: Vandermonde Identity
$(4)$: symmetry of Pascal's Triangle
$(5)$: average $(3)$ and $(4)$
$(6)$: use the binomial theorem to expand $(1+1)^{2n+1}$
$(7)$: simplify

Step $(3)$ is actually an extension of Vandermonde using negative binomial coefficients: $$ \begin{align} \sum_{k=j}^n\binom{k}{j}\binom{2n-k}{n} &=\sum_{k=j}^n\binom{k}{k-j}\binom{2n-k}{n-k}\\ &=\sum_{k=j}^n(-1)^{k-j}\binom{-j-1}{k-j}(-1)^{n-k}\binom{-n-1}{n-k}\\ &=(-1)^{n-j}\binom{-n-j-2}{n-j}\\ &=\binom{2n+1}{n-j}\\ &=\binom{2n+1}{n+j+1} \end{align} $$

Answer 3

With formal power series as requested by OP we may write

$$\sum_{k=0}^n 2^k {2n-k\choose n} = 2^n \sum_{k=0}^n 2^{-k} {n+k\choose n} = 2^n \sum_{k=0}^n 2^{-k} [z^n] (1+z)^{n+k} \\ = 2^n [z^n] (1+z)^n \sum_{k=0}^n 2^{-k} (1+z)^k = 2^n [z^n] (1+z)^n \frac{1-(1+z)^{n+1}/2^{n+1}}{1-(1+z)/2} \\ = 2^{n+1} [z^n] (1+z)^n \frac{1-(1+z)^{n+1}/2^{n+1}}{1-z}.$$

We get from the first piece

$$2^{n+1} [z^n] (1+z)^n \frac{1}{1-z} = 2^{n+1} \sum_{k=0}^n {n\choose k} = 2^{2n+1}.$$

The second piece yields

$$- [z^n] (1+z)^{2n+1} \frac{1}{1-z} = - \sum_{k=0}^n {2n+1\choose k} = - \frac{1}{2} 2^{2n+1}.$$

Joining the two pieces we find

$$2^{2n+1} - 2^{2n} = 2^{2n} = 4^n.$$

Answer 4

Another (shorter) way to approach this is

$$ \bbox[lightyellow] { \eqalign{ & \sum\limits_{0\, \le \,k\, \le \,n} {\left( \matrix{ 2n - k \cr n \cr} \right)2^{\,k} } = \sum\limits_{0\, \le \,k\,\left( { \le \,n} \right)} {\left( \matrix{ 2n - k \cr n - k \cr} \right)2^{\,k} } = \cr & = \sum\limits_{0\, \le \,j\,} {\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{ 2n - k \cr n - k \cr} \right)\left( \matrix{ k \cr k - j \cr} \right)} } = \sum\limits_{0\, \le \,j\,\left( { \le \,n} \right)\,} {\left( \matrix{ 2n + 1 \cr n - j \cr} \right)} = \cr & = {1 \over 2}\sum\limits_{l\,} {\left( \matrix{ 2n + 1 \cr l \cr} \right)} = {{2^{2n + 1} } \over 2} \cr} }$$ where the bounds in brackets means that they are implicit in the binomial, and where in the middle step the Double Convolution is used, which is $$ \bbox[lightyellow] { \eqalign{ & \sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{ a - k \cr n - k \cr} \right)\left( \matrix{ k + b \cr k - m \cr} \right)} = \sum\limits_{\left( {m\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{n - m} \left( \matrix{ n - a - 1 \cr n - k \cr} \right)\left( \matrix{ - m - b - 1 \cr k - m \cr} \right)} = \cr & = \left( { - 1} \right)^{n - m} \left( \matrix{ n - m - a - b - 2 \cr n - m \cr} \right) = \left( \matrix{ a + b + 1 \cr n - m \cr} \right)\quad \left| \matrix{ \;a,b \in R \hfill \cr \;n,m \in Z \hfill \cr} \right. \cr} }$$

Answer 5

Try formal power series harder. The same Snake Oil Method I used to answer one of your other recent questions, works here as well. I will let you work through this one on your own with a few hints. Let $B=B(x)=\frac{1}{\sqrt{1-4x}}$ and $C=C(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating functions for the central binomial coefficients and Catalan numbers, respectively. Then $$ B=\frac{1}{1-2xC}, \qquad B^2=\frac{1}{1-4x}, $$ and $$ \sum_{n=0}^{\infty}\binom{2n+k}{n}x^n=BC^k. $$ The last inequality can be proved combinatorially: consider all lattice paths from $(0,0)$ to $(2n,k)$ using steps $u=(1,1)$ and $d=(1,-1)$. Then any such path can be written uniquely as $P_0uD_1uD_2u\dots uD_k$, where $P_0$ is a grand Dyck path and $D_1,\dots,D_k$ are Dyck paths, all together (with $P_0$) of total semilength $n$.