Evaluation of a series (possibly related to Binomial Theorem)

I have the following series:

$$1 + \frac{2}{3}\cdot\frac{1}{2} + \frac{2\cdot5}{3\cdot6}\cdot\frac{1}{2^2} + \frac{2\cdot5\cdot8}{3\cdot6\cdot9}\cdot\frac{1}{2^3} + \ldots$$

I have to find the value of this series, and I have four options:
(A) $2^{1/3}$ (B) $2^{2/3}$ (C) $3^{1/2}$ (D) $3^{3/2}$

I can't seem to find a general term for this. I tried:

$$S = 1 + \frac{(1 - \frac{1}{3})}{1!}(\frac{1}{2}) + \frac{(1 - \frac{1}{3})(2 - \frac{1}{3})}{2!}(\frac{1}{2})^2 + \frac{(1 - \frac{1}{3})(2 - \frac{1}{3})(3 - \frac{1}{3})}{3!}(\frac{1}{2})^3 + \ldots$$

But this doesn't seem to get me anywhere.

Any help?

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This maybe a telescopic series, because there was a similar question we solved in class which ended up being telescopic:

$$ \frac{3}{2^3} + \frac{4}{2^4\cdot3} + \frac{5}{2^6\cdot3} + \frac{6}{2^7\cdot5} + \ldots$$

$=\displaystyle\sum\limits_{r=1}^\infty\frac{r+2}{2^{r+1}r(r+1)}$

$=\displaystyle\sum \bigg(\frac{1}{2^r r} - \frac{1}{2^{r+1}(r+1)}\bigg) = \frac{1}{2}$

$P.S:$ This problem was included in my set of questions for Binomial Theorem, which is why I thought it might be related to it.