Binomial type sum Does the following reduce to something simpler
$\sum_j {k \choose 2j} x^{2j}$
I have perused the combinatorial identities by could but I did not find anything that fits my case.

HINT
we say $\sum_j{k\choose 2j} a_j$ to be the $even $ $aerated $ $binomial $ $transformation$ of the sequence ${a_j}$. Now if $b_j$ is another sequence such that $∆b_j$$=a_j$ for $j>0$
we can easily show that
THEOREM $1$
if ${g_k}$ is the binomial transform of ${a_j}$ $∆g_k-a_0[n=-1]$ is the binomial transform of $a_{j+1}$, binomial transform is defined by $g_k=\sum_j{k\choose j} a_j $
Let $G_k$ and $H_k$ be the even aerated binomoial transforms of $a_j$ and $b_j$ respectively, then for all $k$ consider the sequence {$b_0,0,b_1,0,b_2,....$}, theorem $1$ and get the new sequence {$b_1,0,b_2,....$}. the binomial transform of  {$b_1,0,b_2,....$} is given by {$H_{k+2}-2H_{k+1}+H_n+b_0[k=-1]-b_0[k=-2]$}.Since $∆b_j$$=a_j$ for $j>0$ so for all $k$ $G_k=H_{k+2}-2H_{k+1}+H_n+b_0[k=-1]-b_0[k=-2]-H_n$=$H_{k+2}-2H_{k+1}+b_0[k=-1]-b_0[k=-2]$
I wish this helps you.

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*if $P$ is a statement $[P]$ is $1$ for $P$ is true and $0$ otherwise.


*$∆b_j=b_{j+1}-b_j$
