Let $b\ge 1$ and $1 \le q \le 2$, can we show that there exists an $0\le a \le b$ and $p\in [0,1]$ such that
\begin{align} b^{2n} = (1-p) + p \sum_{k=0}^n c(n,k,q) a^{2k} \end{align} for every integer $n\ge 1$ where the coefficients in the series have the following complicated expression \begin{align} c(n,k,q)=\frac{ \mathrm{B} \left( \frac{2k+1}{q}, \frac{2n-2k+1}{q}\right) }{ (2n+1) \mathrm{B}( \frac{2n+1}{q},\frac{1}{q} ) \mathrm{B} \left( 2k+1, 2n-2k+1\right) } \end{align}
where the beta function is defined as $\mathrm{B} \left( x, y\right)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$.
Solution for $q=1$. If $q=1$ then it is not difficult to check that \begin{align} c(n,k,q)=1 \end{align}
and we have to solve
\begin{align} b^{2n} = (1-p) + p \sum_{k=0}^n a^{2k} \end{align} using the sum $\sum_{k=0}^n a^{2k} =\frac{a^{2k+1}-1}{a^2-1}$ we have to solve
\begin{align} b^{2n} = (1-p) + p \frac{a^{2k+1}-1}{a^2-1} \end{align} which is done by using $1-p=\frac{1}{b^2}$ and $a=b$.
Solution for $q=1$. For $q=2$ it is more tedious but one can check that the solution is given by $p=1$ and $a^2=b^2-1$.
So, the question is can we show this for $q \in (1,2)$. Note that we are not interested in what the solution is but rather that the solution esits.
There exist no such $a,p$ in general. The following will derive $a,p$ from the equations for $n=1,2$, then substitute them for $n=3$ and show that the equality fails for $q=\frac{3}{2}$.
First note that $c(n,k,q)=c(n,n-k,q)$ and $c(n,0,q)=1$. Let $c=c(2,1,q)$ then:
$$ \require{cancel} \begin{cases} b^2 & = 1-\cancel{p}+p(\cancel{1}+a^2) & = 1 + pa^2 \\ b^4 & = 1 -\cancel{p} + p(\cancel{1}+a^4) + pc a^2 & = 1+pca^2+pa^4 \begin{align} \end{align} \end{cases} $$
Eliminating $p$ between the equations gives $a^2$ then substituting back gives $p$. In the end:
$$ \begin{cases} \begin{align} a^2 & = b^2 + 1 - c \\ p & = \frac{b^2-1}{b^2+1-c} \end{align} \end{cases} $$
Let $d=c(3,1,q)=c(3,2,q)$ then the equation for $n=3$ is:
$$ b^6 = 1 - \cancel{p} + p(\cancel{1}+a^6) + pd(a^2+a^4) = 1+pa^2(a^4+da^2+d) $$
Substituting $a,p$ as determined at the previous step, and after simple manipulations:
$$ (b^2+1-c)(b^4+b^2+1) = (b^2+1-c)\big((b^2+1-c)^2+d(b^2+1-c)+d\big) $$
Since this must hold as a polynomial identity in $b$, the coefficients must match between the two sides. In particular, equating the free terms gives:
$$ 1-c = (1-c)\big((1-c)^2+d(1-c) + d\big) $$
So either $c=1$ or else:
$$ \begin{align} (1-c)^2+d(1-c) + d = 1 \iff \cancel{1}-2c+c^2+d(2-c)=\cancel{1} \iff (c-2)(c-d)=0 \end{align} $$
Therefore a necessary condition for the expansion to hold is for either of the following to be true:
$$ \begin{cases} c(2,1,q)=1 & \quad\quad \text{(true for } q=1 \text{)}\\ c(2,1,q)=2 & \quad\quad \text{(true for } q=2 \text{)} \\ c(3,1,q)=c(2,1,q) \end{cases} $$
For $q=\frac{3}{2}$, however, none of the conditions hold true, therefore the expansion fails at $n=3$:
$$ \begin{cases} \begin{align} c(2,1,\frac{3}{2}) & = \frac{6}{\Gamma\left(\frac{2}{3}\right) \Gamma\left(\frac{10}{3}\right)} \simeq 1.59492 \\ c(3,1, \frac{3}{2}) & = \frac{27 \,\Gamma\left(\frac{16}{3}\right)}{26 \,\Gamma\left(\frac{2}{3}\right) \Gamma\left(\frac{14}{3}\right)} \simeq 2.09188 \end{align} \end{cases} $$