Summing a series having a geometric component 
If $$\alpha=\frac {5}{2!3}+\frac {5\cdot 7}{3!3^2}+\frac {5\cdot 7\cdot 9}{4!3^3}+\dots $$  Find the value of $\alpha^2+4\alpha $.

The possible options are:


*

*21

*23

*25

*27
I think $$\alpha=\sum \frac {(2n+4)!}{2^{n+2}(n+1)!(n+2)!3^{n+1}}$$ but cannot think of what else to do.
 A: Consider the binomial expansion of $$(1-\frac 23)^{-\frac 32}$$
This has value $3\sqrt{3}$ and when expanded gives the sequence $$1+1+\frac{5}{2!3}+\frac{5\cdot7}{3!3^2}+...$$
So $$\alpha=3\sqrt{3}-2$$ which gives the required expression the value of $23$
A: Indeed
$$\alpha=\dfrac16\sum_{n=2}^\infty\dfrac{(2n+1)!}{(n!)^2}\left(\dfrac16\right)^n$$
one may construct a binomial series and show that with the generating function
$$\alpha(x)=-\dfrac16-x+\dfrac16\dfrac{1}{\sqrt{1-4x}^3}$$
then $\alpha=\alpha(\dfrac16)$.
A: So we have that
$$\alpha=\sum_{n=0} \frac{(5+2n)!!}{3^{n+2}(2+n)!}$$
Using the fact that $(2n)!=2^n(2n-1)!!(n!)$, we have that
$$\alpha=\sum_{n=0} \frac{(2n+6)!}{2^{n+3}3^{n+2}(n+2)!(n+3)!}$$
$$\alpha=\frac{1}{2}\sum_{n=0} \binom{2n+6}{n+3}\left(\frac{1}{6}\right)^{n+2}(n+3)$$
Using the generating function of central binomial coefficients, we know
$$\sum_{n=0}^\infty \binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$$
Define the function
$$f(x)=\sum_{n=0}^{\infty}\binom{2n+6}{n+3}x^{n+3}=\frac{1}{\sqrt{1-4x}}-1-2x-6x^2$$
We know
$$f'(x)=\sum_{n=0}^\infty \binom{2n+6}{n+3}x^{n+2}(n+3)=\frac{2}{\left(\sqrt{1-4x}\right)^3}-2-12x$$
Hence,
$$\alpha=\frac{1}{2}f'\left(\frac{1}{6}\right)=3\sqrt{3}-2$$
We can then deduce that $\alpha$ is a root of $(x+2)^2=27$, or $x^2+4x-23=0$. Hence,  $\alpha^2+4\alpha=\boxed{23}$
