finding a relation in $p:p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$ 
if $$p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$$ and $$p^2+ap+c=0.$$ Find $a,c$ also $|c|=2$

My progress:The general term $$T_{m+1}=\frac{(1)(3)\cdots(2m+1)}{(3)(6)\cdots(3m+3)}$$ from here i tried to make it in terms of factorial but it does not help.
Is there any algorithm to solve these problems?
(This question came in a maths magazine

Since $p=\tfrac13\sum_{m\ge0}\tfrac{\binom{2m+1}{m}}{6^m}$ and $\binom{n}{k}=\oint_{|z|=1}\frac{(1+z)^ndz}{2\pi iz^{k+1}}$, simplifying a geometric-series integrand eventually gives$$p=-2\oint_{|z|=1}\frac{1+z}{1-4z+z^2}\frac{dz}{2\pi i}.$$The denominator's roots are $2\pm\sqrt{3}$, but only $2-\sqrt{3}$ is enclosed. So$$p=-2\lim_{z\to2-\sqrt{3}}\frac{1+z}{z-2-\sqrt{3}}=\sqrt{3}-1.$$(More generally, the above technique shows $\sum_{m\ge0}\binom{2m+1}{m}x^m=\frac{(1-4x)^{-1/2}-1}{2x}$ for $|x|<1/4$.) If we assume $a,\,c\in\Bbb Z$, the sought quadratic has roots $-1\pm\sqrt{3}$ so that $a=2,\,c=-2$ because the quadratic is $(p+1)^2-3=0$.
