Binomial type sum

Question

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.

Answer 1

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.

• $∆b_j=b_{j+1}-b_j$

Answer 2

\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}