Show that: $1+\dfrac {1}{4}+\dfrac {1\cdot 4}{4\cdot 8}+\dfrac {1\cdot4\cdot7}{4\cdot8\cdot12}+ \dotsb = (2)^{2/3}$

Question

Show that: $1+\dfrac {1}{4}+\dfrac {1\cdot4}{4\cdot8}+\dfrac {1\cdot4\cdot7}{4\cdot8\cdot12}+ \dotsb= (2)^{2/3}$

My attempt:

$$R.H.S=(2)^{\dfrac {2}{3}}$$ $$=(3-1)^{\dfrac {2}{3}}$$ $$=3^{\dfrac {2}{3}} [1+(-\dfrac {1}{3})]^{\dfrac {2}{3}}$$ $$=3^{\dfrac {2}{3}} [1+\dfrac {2}{3} \times \dfrac {-1}{3} + \dfrac {2}{3} \times \dfrac {\dfrac {2}{3} -1}{2!} \times (\dfrac {-1}{3})^2 +....$$

This doesn't seem to give the required resullt.

Answer 1

Expand $$\left(1-\frac34\right)^{-1/3}$$ by the binomial theorem. One gets $$4^{1/3}=\sum_{n=0}^\infty\frac{1\cdot4\cdots(3n-2)}{4\cdot8\cdots(4n)}.$$