closed form of $\sum_{k=0}^n {2n\choose 2k}2^k$

Question

Is it possible to find a closed form for the expression below? $$\sum_{k=0}^n {2n\choose 2k}2^k$$

I have tried counting in two ways but made no progress.

And I don't know any combinatorial identities that help me simplify the sigma.

Answer 1

Hint: $$\dfrac{1}{2} \sum_{j=0}^{2n} {2n \choose j} 2^{j/2} + \dfrac{1}{2} \sum_{j=0}^{2n} {2n \choose j} (-2)^{j/2}$$

Answer 2

$$A_n=\sum_{k=0}^{n}\binom{2n}{2k}2^k = \sum_{k=0}^{n}\binom{2n}{2k}\sqrt{2}^{2k} = \frac{1}{2}\sum_{j=0}^{n}\binom{2n}{j}\sqrt{2}^j\left(1^j+(-1)^j\right) $$ hence $A_n$ is related with Pell-Lucas numbers:

$$ A_n = \frac{(1+\sqrt{2})^{2n}+(1-\sqrt{2})^{2n}}{2}.$$

To prove the same, you may also check that $A_{n+2}=6\,A_{n+1}-A_n$ and $A_0=1, A_1=3$.

A combinatorial interpretation is the following: you have $2n$ officials and you want to first promote to a higher ranking an even number of them, then promote again at most half of the previously promoted officials.