Computing the binomial series: $\binom{n}{2}\cdot2^2+\binom{n}{4}\cdot2^4+\binom{n}{6}\cdot2^6+\cdots+ \binom{n}{n}\cdot2^n$ 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.
 A: $$3^n = (1+2)^n=1+\binom{n}{1}*2^1+\binom{n}{2}*2^2+\binom{n}{3}*2^3\ldots,+ \binom{n}{n}*2^{n}$$
$$ (-1)^n =(1-2)^n=1-\binom{n}{1}*2^1+\binom{n}{2}*2^2-\binom{n}{3}*2^3\ldots,+ \binom{n}{n}*2^{n}$$
then $$3^n +(-1)^n = (1+2)^n+(1-2)^n=2+\binom{n}{2}*2^2+\binom{n}{4}*2^4+\binom{n}{6}*2^6\ldots, $$
A: 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?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
\sum_{k = 1}^{\infty}{n \choose 2k}2^{2k} & =
\sum_{k = 2}^{\infty}{n \choose k}2^{k}\,{1 + \pars{-1}^{k} \over 2} =
{\sum_{k = 2}^{\infty}{n \choose k}2^{k} +
\sum_{k = 2}^{\infty}{n \choose k}\pars{-2}^{k} \over 2}
\\[5mm] & =
{\bracks{\pars{1 + 2}^{\, n} - 1 - 2n} +
\braces{\vphantom{\large A}\bracks{1 + \pars{-2}}^{\, n} - 1 + 2n}
\over 2}
\\[5mm] & =
\bbx{3^{n} + \pars{-1}^{n} - 2 \over 2}
\end{align}
