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

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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}
