Then value of $\alpha^2 +4\alpha$ in Infinite series 
If $\displaystyle \alpha = \frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots \infty.$
Then value of $\alpha^2 +4\alpha$ is

Try: Let $$S = \frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots $$
$$S+1 = 1+\frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots $$
Now camparing with $$(1+x)^n = 1+nx+\frac{n(n-1)x^2}{2!}+\frac{n(n-1)(n-2)x^3}{6\cdot 3!}+\cdots \cdots$$
So $\displaystyle nx=\frac{5}{6}$ and $\displaystyle \frac{n(n-1)x^2}{2}=\frac{35}{27}$
So $$\frac{nx(nx-x)}{2}=\frac{5}{12}\cdot \frac{5-6x}{6}=\frac{35}{27}$$
So $\displaystyle x=-\frac{41}{18}$ and $\displaystyle n=-\frac{15}{41}$
I am getting $\displaystyle S+1=\bigg(1-\frac{41}{18}\bigg)^{-\frac{15}{41}}$
but answer of $\alpha^2+4\alpha = 23$
which is not possible from my answer. could some help me how can i solve it, thanks
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Indeed, $\ds{\alpha}$ is given by
\begin{align}
\alpha & \equiv
\sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{2k + 3} \over
\pars{n + 1}!\,3^{n}} =
\sum_{n = 1}^{\infty}{2^{n}\prod_{k = 1}^{n}\pars{k + 3/2} \over
\pars{n + 1}!\,3^{n}}
\\[5mm] & =
\sum_{n = 1}^{\infty}{\pars{5/2}^{\large \overline{n}} \over
\pars{n + 1}!}\pars{2 \over 3}^{n} =
\sum_{n = 1}^{\infty}{\Gamma\pars{5/2 + n}/\Gamma\pars{5/2} \over
\pars{n + 1}!}\pars{2 \over 3}^{n}
\\[5mm] & =
\sum_{n = 1}^{\infty}{\pars{n + 3/2}! \over
\pars{n + 1}!\pars{3/2}!}\pars{2 \over 3}^{n} =
\sum_{n = 1}^{\infty}{1 \over n + 1}{n + 3/2 \choose n}
\pars{2 \over 3}^{n}
\\[5mm] & =
\sum_{n = 1}^{\infty}\pars{\int_{0}^{1}t^{n}\,\dd t}
\bracks{{-5/2 \choose n}\pars{-1}^{n}}
\pars{2 \over 3}^{n}
\\[5mm] & =
\int_{0}^{1}\sum_{n = 1}^{\infty}{-5/2 \choose n}
\pars{-\,{2 \over 3}\,t}^{n}\,\dd t =
\int_{0}^{1}\bracks{\pars{1 - {2 \over 3}\,t}^{-5/2} - 1}\,\dd t
\\[5mm] &
\implies \bbx{\alpha = 3\root{3} - 2} \implies
\bbx{\alpha^{2} + 4\alpha = \color{red}{\large 23}}
\end{align}
A: Using 
\begin{align}
\prod_{k=1}^{n} (2 k +3) &= 2^{n} \, \prod_{k=1}^{n} \left(k + \frac{3}{2}\right) = \frac{2^{n} \, \left(n + \frac{3}{2}\right)!}{\left(\frac{3}{2}\right)!} \\
(a)_{n} &= \frac{\Gamma(n+a)}{\Gamma(a)} \\
\sum_{n=0}^{\infty} \frac{(a)_{n} \, x^n}{n!} &= (1-x)^{-a}
\end{align}
then the following is determined:
\begin{align}
\alpha &= \sum_{k=1}^{\infty} \frac{\prod_{s=1}^{k} (2s + 3) }{(k+1)! \, 3^{k}} \\
&= \frac{1}{\Gamma(5/2)} \, \sum_{k=1}^{\infty} \frac{\Gamma\left(k + \frac{5}{2}\right) }{(k+1)!} \, \left(\frac{2}{3}\right)^{k} \\
&= \frac{3}{2} \, \frac{\Gamma(3/2)}{\Gamma(5/2)} \, \sum_{k=2}^{\infty} \frac{(3/2)_{k}}{k!} \, \left(\frac{2}{3}\right)^{k} \\
&= \left(1 - \frac{2}{3}\right)^{-3/2} - 1 - 1 \\
&= 3^{3/2} - 2.
\end{align}
Now,
$$\alpha^{2} + 4 \alpha = \alpha (\alpha + 4) = (3^{3/2} -2)(3^{3/2}+2) = 3^{3} - 2^{2}.$$
