$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24} + \cdots.$ How to find the sum 

$$\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24}  + \cdots.$$

I have tried to make this series in  the form $1 + n x  + \frac {n (n + 1) x^2}{2!} + \frac {n (n+1) (n+2)x^3}{3!} +  \cdots$. But I failed to do so. Can anyone please help me

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? $,
If $\displaystyle S=\frac {1}{6} + \frac {5}{6.12} + \frac {5. 8}{6.12.18} + \frac {5.8.11}{6.12.18.24}  + \cdots.$
$$2S=\frac {2}{6} + \frac {2\cdot5}{6.12} + \frac {2\cdot5. 8}{6.12.18} + \frac {2\cdot5.8.11}{6.12.18.24}  + \cdots$$
$$2S+1+\dfrac13$$
$$=1+\dfrac{-\dfrac23}{1!}\left(-\dfrac36\right)+\dfrac{-\dfrac23\left(-\dfrac23-1\right)}{2!}\left(-\dfrac36\right)^2+\dfrac{-\dfrac23\left(-\dfrac23-1\right)\left(-\dfrac23-2\right)}{3!}\left(-\dfrac36\right)^3+ \cdots$$
$$=\left(1-\dfrac12\right)^{-2/3}$$
