Computing the binomial series: $\binom{n}{2}\cdot2^2+\binom{n}{4}\cdot2^4+\binom{n}{6}\cdot2^6+\cdots+ \binom{n}{n}\cdot2^n$

Question

How to find the tight bound for the binomial series $\binom{n}{2}\cdot2^2+\binom{n}{4}\cdot2^4+\binom{n}{6}\cdot2^6\ldots,+ \binom{n}{n}\cdot2^n$

I have found $3^n$ for series $(1+2)^n=1+\binom{n}{1}\cdot2^1+\binom{n}{2}\cdot2^2+\binom{n}{3}\cdot2^3+\cdots+ \binom{n}{n}\cdot2^n$

Is there any way to prove the bound of above series.

Answer 1

Hint:

$$(a+b)^n+(a-b)^n=2\sum_{r=0}^{n/2}\binom n{2r} a^{n-2r}b^{2r}$$

Can you recognize $a,b$ here?

Answer 2

$$3^n = (1+2)^n=1+\binom{n}{1}*2^1+\binom{n}{2}*2^2+\binom{n}{3}*2^3\ldots,+ \binom{n}{n}*2^{n}$$

$$ (-1)^n =(1-2)^n=1-\binom{n}{1}*2^1+\binom{n}{2}*2^2-\binom{n}{3}*2^3\ldots,+ \binom{n}{n}*2^{n}$$ then $$3^n +(-1)^n = (1+2)^n+(1-2)^n=2+\binom{n}{2}*2^2+\binom{n}{4}*2^4+\binom{n}{6}*2^6\ldots, $$