Let $b\ge 1$, can we show that there exists an $0\le a $ and $p\in [0,1]$ such that
\begin{align} b^{2n} =(1-p)+p \frac{1}{2n+1} \frac{a^{2n+2}-1}{a^2-1}, \end{align} for every posive integer $n$.
Note, that I am not interested in the solution only in the existence of the solution.
Note that for each $n$ this can easily be done. However, I want the statement to hold for all $n$.
It is not possible for the equality to hold for $\forall n \in \mathbb{N}$ except for the trivial case $b=1$ where you can choose $p=0$.
If the equality were to hold for $b \gt 1$, then at the limit:
$$ \begin{align} 1 & = \lim_{n \to \infty} \cfrac{(1-p)+p \frac{1}{2n+1} \frac{a^{2n+2}-1}{a^2-1}}{b^{2n}} \\ & = \lim_{n \to \infty} \cfrac{1-p}{b^{2n}} - \lim_{n \to \infty} \cfrac{p}{(2n+1)(a^2-1)b^{2n}} + \lim_{n \to \infty} \cfrac{p\,a^{2n+2}}{(2n+1)(a^2-1)b^{2n}} \\ & = 0 - 0 + \cfrac{p\,a^2}{a^2-1}\,\lim_{n \to \infty} \cfrac{\left(\frac{a}{b}\right)^{2n}}{2n+1} \end{align} $$
The latter limit is $0$ if $a \le b$ and $\infty$ if $a \gt b$, so the resulting limit can not be $1$ regardless of $p,a$.