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
 A: $p=\frac{1}{3}+\frac{1\cdot 3}{1 \cdot 2}\cdot (\frac{1}{3})^2+\frac{1\cdot 3\cdot 5}{1 \cdot 2\cdot 3}\cdot (\frac{1}{3})^3 + \cdots$
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?
A: It turns out that $p = \sqrt{3} - 1$. To see this, you need to recognize $p$ as the power series of the function $1/\sqrt{1-x} - 1$, evaluated at $2/3$. More specifically, the $n$-th derivative of $(1-x)^{-1/2}$ is
$$
\begin{align}
\frac{1}{2}\cdot \frac{3}{2}\cdots \frac{2n-1}{2} (1-x)^{-\frac{1}{2} - n} & = (1-x)^{-\frac{1}{2} - n}  \prod_{i=0}^{n-1}(-1)\left(-\frac{1}{2}-i\right)\\
& = (1-x)^{-\frac{1}{2} - n}  \prod_{i=1}^{n}\frac{2i-1}{2}
\end{align}
$$
Evaluating $ (1-x)^{-\frac{1}{2} - n} $ at zero just gives us $1$ for all $n$, so the power series should be
$$
1 + \frac{1}{2} x + \frac{1}{2!} \left(\frac{1}{2}\cdot\frac{3}{2}\right) x^2 + \frac{1}{3!}\left(\frac{1}{2}\cdot \frac{3}{2}\cdot\frac{5}{2} \right)x^3 + \cdots
$$
or
$$
\frac{1}{\sqrt{1-x}} = \sum_{n = 0}^\infty \frac{x^n}{n!} \prod_{i = 1}^{n} \frac{2i-1}{2} = \sum_{n=0}^\infty \prod_{i=1}^n\frac{2i-1}{2i} x
$$
Plugging in $x = 2/3$ then gives us
$$
\frac{1}{\sqrt{1-2/3}} = \sqrt{3} = \sum_{n=0}^\infty \prod_{i=1}^{n} \frac{2i-1}{2 i} \frac{2}{3} = \sum_{n=0}^\infty \prod_{i=1}^n \frac{2i-1}{3i}\ .
$$
The right hand side can now be recognized as $p+1$.
Finally, to construct a quadratic equation that has $p$ as one of its roots, you can make use of the Vieta's formulas. Under the constraint $|c| = 2$, there are really only two possibilities. If $c = 2$, then the other root $p'$ should be $c/p = 1+\sqrt{3}$, which tells us that $a = -p-p' = 2\sqrt{3}$. The $c = -2$ case is similar and I'll leave it to you to verify that $a = 2$ there.
The key recognition is inspired by the form of the general term. Its denominator contains a factorial, and its numerator contains a product of an arithmetic progression. The former suggests that we might be able to rewrite $p$ as a power series and the latter suggests that it might be the series of some negative power of $x$. The lack of alternating signs then suggests that we need to compose it with $1-x$ to get an extra minus sign each time we take the derivative.
A: 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 $(p+1)^2-3=0$.
