Finding the summation of a series Determine the following sum
$$\frac{5}{3 \cdot 6 }\cdot \frac{1}{4^2} + \frac{5\cdot 8}{3 \cdot 6 \cdot 9 }\cdot \frac{1}{4^3} + \frac{5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\cdot \frac{1}{4^4} + \frac{5\cdot8\cdot 11\cdot 14}{3 \cdot 6\cdot 9\cdot 12 \cdot 15}\cdot \frac{1}{4^5} + \dots \dots$$
I tried to use Generalised Binomial Theorem, but I am unable to find $x,y,r$

The Generalized Binomial Theorem
  $$(x+y)^{r}=\sum _{k=0}^{\infty }{r \choose k}x^{r-k}y^{k}$$ where the ${r \choose k }$ denotes the falling factorial.
  I notice that the factors in the numerator is increasing, hence I am unable to use it.

 A: Hint: $$\frac{5 \cdot 8 \cdot 11 \cdot 14}{3 \cdot 6 \cdot 9 \cdot 12 \cdot 15}=\frac{1}{3 \cdot 5!}(1+\frac{2}{3})(2+\frac{2}{3})(3+\frac{2}{3})(4+\frac{2}{3})=\frac{(-1)^4}{3 \cdot 5!}\prod_{k=0}^3{\left(-\frac{5}{3}-k\right)}$$. 
A: As the denominator has one extra multiplicand 
and the number of terms in denominator $=$ the exponent of $\dfrac14$ 
$$2S=\dfrac{2\cdot5}{3\cdot6}\left(\dfrac14\right)^2+\dfrac{2\cdot5\cdot8}{3\cdot6\cdot9}\left(\dfrac14\right)^3+\cdots$$
Now like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
$$2S+1=1+\dfrac{\dfrac23\cdot\dfrac53}{2!}\left(\dfrac14\right)^2+\cdots$$
$$=1+\dfrac{\left(-\dfrac23\right)\cdot\left(-\dfrac23-1\right)} 
 {2!}\left(-\dfrac14\right)^2+\cdots$$
$$=\left(1-\dfrac14\right)^{-2/3}$$  using Binomial series
We can use Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $ 
to find  $x,n$ to be $-\dfrac14-\dfrac23$ respectively in $$2S+1=(1+x)^n$$
