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

$p=\frac{1}{3}+\frac{1\cdot 3}{1 \cdot 2}\cdot (\frac{1}{3})^2+\frac{1\cdot 3\cdot 5}{1 \cdot 2\cdot 3}\cdot (\frac{1}{3})^3 + \cdots$
So $p+1=1+\frac{1}{3}+\frac{1\cdot 3}{1 \cdot 2}\cdot (\frac{1}{3})^2+\frac{1\cdot 3\cdot 5}{1 \cdot 2\cdot 3}\cdot (\frac{1}{3})^3 +\cdots$
Now notice that $$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots$$
We can assume that $$nx=\frac{1}{3}$$ and $$\frac{n(n-1)}{2!}x^2=\frac{1\cdot 3}{1 \cdot 2}\cdot (\frac{1}{3})^2$$
and we will get $$\frac{n-1}{n}=3\Longrightarrow n=-\frac{1}{2}\Longrightarrow x=-\frac{2}{3}$$
Therefore we have $$1+p=\frac{1}{\sqrt{1+x}}=\sqrt{3}$$ Can you end it now?
