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.
 A: \begin{align}
\sum_{j \ge 0} \binom{k}{2j} x^{2j}
&= \sum_{j \ge 0} \frac{1+(-1)^j}{2}\binom{k}{j} x^j \\
&= \frac{1}{2}\sum_{j \ge 0} \binom{k}{j} x^j + \frac{1}{2}\sum_{j \ge 0} \binom{k}{j} (-x)^j \\
&= \frac{(1+x)^k + (1-x)^k}{2} 
\end{align}
A: 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.

*

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


*$∆b_j=b_{j+1}-b_j$
