I am trying to find the positive real number $k$ such that: $$(k+1)+\sum^{\infty}_{n=1} \left(\frac{\binom{4n}{n}}{3n+1} k^{2n}+\frac{2\binom{4n+1}{n}}{3n+2} k^{2n+1} \right)=2$$
This problem arose from the following question: Consider the sequence of real numbers $(a_n)$ with $a_0=1$ and for every $n \geq 1$, $a_n=a_{n-1}a_0+a_{n-3}a_2+\cdots+a_0 a_{n-1}$ if $n$ is odd and $a_n=a_{n-1}a_1+a_{n-3}a_3+\cdots+a_1 a_{n-1}$ if $n$ is even. Compute the positive real number $k$ such that $\sum^{\infty}_{i=0} a_i k^i=2$.
I began by computing the terms of the sequence up to $a_{11}$, which yields $a_0=a_1=a_2=1$, $a_3=2$, $a_4=4$, $a_5=9$, $a_6=22$, $a_7=52$, $a_8=140$, $a_9=340$, $a_{10}=969$, and $a_{11}=2394$. Considering $a_n$ when $n$ is odd and when $n$ is even and searching in the OEIS database led me to conjecture that $a_{2n}=\frac{\binom{4n}{n}}{3n+1}$ for integers $n \geq 1$ and $a_{2n+1}=\frac{2\binom{4n+1}{n}}{3n+2}$ for integers $n \geq 1$.
I'm fairly sure this is correct because inputting the sum into Wolfram Alpha yields a $-(k+1)$ in the closed form of the sum mentioned at the start, which cancels with $a_0+a_1k$.
However, the closed form involved the hypergeometric function, which I don't think is the intended solution since this problem should be solvable by high schoolers.
