TITLE: Sum of infinite series $ {1+ \frac{2}{6} + \frac{2\cdot5}{6\cdot12} + \frac{2\cdot5\cdot8}{6\cdot12\cdot18} + \cdots}$.
QUESTION [3 upvotes]: Prove that $1+ \frac{2}{6} + \frac{2\cdot5}{6\cdot12} + \frac{2\cdot5\cdot8}{6\cdot12\cdot18} +\cdots=4^{\frac13}$
I tried it in the backward method... I rewrote $4^{\frac13}$ in this way... 
$(1+3)^{\frac13}$ and expanded it in the binomial expansion method, but it doesn't help in any way.

REPLY [3 votes]: using binomial expansion $(1-x)^{-\frac{2}{3}} =1-\frac{2}{3}(-x)+\frac{\frac{-2}{3}\times\frac{-5}{3}}{2!}(-x)^2+\frac{\frac{-2}{3} \cdot \frac{-5}{3} \cdot \frac{-8}{3}}{3!} \cdot (-x)^3 +\ldots$
when $x=\frac{1}{2}$ we get: 
$(1-\frac{1}{2})^{-\frac{2}{3}}=(\frac{1}{2})^{-\frac{2}{3}}=(2)^\frac{2}{3}=(2^2)^\frac{1}{3}=4^{\frac{1}{3}}$
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