The sums $\sum_{0\leq k}\binom{n}{2k}x^k$ and $\sum_{0\leq k}\binom{n}{2k+1}x^k$

Question

Let $n$ be a positive integer and suppose that $$ \sum_{0\leq k}\binom{n}{2k}x^k=A $$ What can we say about the sum $$ \sum_{0\leq k}\binom{n}{2k+1}x^k\ ? $$ Thanks!

Edit: Can we express the second sum using the value A of the first sum ?

Answer 1

Consider the equation $$ \sum_{0\leq k}\binom{n}{2k}x^k-A=0. $$ Since this is a polynomial equation of order $m=\left\lfloor\frac n2\right\rfloor$, it has (taking into account the multiplicity) exactly $m$ roots $x_i(A)$, parametrically dependent on $A$, which can be easily found numerically. The corresponding $B$ can be then calculated as $$ B_i(A)=\sum_{0\leq k}\binom{n}{2k+1}x_i^k(A). $$

The numerical calculations do not support a speculation that there can be a general simple formula to express all $B_i$ (or even one of them) in terms of $A$. This however does not exclude a simple expression for some specific values of $A$. For example if $A=1$ one of the roots is $0$ and the corresponding $B$ is $n$.