Is there a closed form for $\sum_{k=0}^n2^k\binom{2n+1}{2k}$, the sum of binomial coefficients times powers of two, for even indices?

Question

I'm hoping for a nice and simple closed form for the sum $$\sum_{k=0}^{n}2^k\binom{2n+1}{2k}.$$ Searching this site I found many nondescript titles but no duplicates, though I wouldn't be surprised if this has been asked before.

EDIT: Both answers answer my question perfectly, and made me realise that I asked the wrong question. So if possible, I'm looking for a closed form in terms of only $n$ and rational numbers, if that makes sense.

Answer 1

Write $\mathbf{1}_{\{p\}}$ for the function which takes values $1$ when $p$ is true and $0$ when $p$ is false. Then

\begin{align*} \sum_{k=0}^{n} \binom{2n+1}{2k} x^k &= \sum_{j=0}^{2n+1} \binom{2n+1}{j} x^{j/2} \mathbf{1}_{\{\text{$j$ is even}\}} \\ &= \sum_{j=0}^{2n+1} \binom{2n+1}{j} x^{j/2} \cdot \frac{1 + (-1)^j}{2} \\ &= \frac{(1+x^{1/2})^{2n+1} + (1-x^{1/2})^{2n+1}}{2}. \end{align*}

Here is a good exercise for this line of reasoning:

Exercise. Let $\omega = \frac{-1+i\sqrt{3}}{2}$. By using the fact that $(1 + \omega^k + \omega^{2k})/3 = \mathbf{1}_{\{k \equiv 0 \pmod{3}\}}$, prove that

$$ \sum_{k=0}^{\lfloor n/3 \rfloor} \binom{n}{3k} x^{3k} = \frac{(1+x)^n + (1+\omega x)^n + (1 + \omega^2 x)^n}{3} $$

In general, let $\omega$ a primitive $m$-th root of $1$ in $\mathbb{C}$, meaning that $m$ is the smallest positive integer for which $\omega^m = 1$ holds.

(1) Prove that $\frac{1}{m}\sum_{l=0}^{m-1} \omega^{kl} = \mathbf{1}_{\{ k \equiv 0 \pmod{m}\}}$.

(2) Let $r \in \{0,\cdots,m-1\}$. Prove that

$$\sum_{k=0}^{\lfloor (n-r)/m \rfloor} \binom{n}{mk+r} x^{mk+r} = \frac{1}{m} \sum_{l=0}^{m-1} \omega^{-rl}(1+\omega^l x)^n . $$