<s>[INST] 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.
 [/INST] 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 $23$
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