# Sum of infinite series ${1+ \frac{2}{6} + \frac{2\cdot5}{6\cdot12} + \frac{2\cdot5\cdot8}{6\cdot12\cdot18} + \cdots}$.

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.

• Consider the binomial series for $(1-x)^{-2/3}$. – Lord Shark the Unknown Jun 27 '18 at 18:09

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:
$(1-\frac{1}{2})^{-\frac{2}{3}}=(\frac{1}{2})^{-\frac{2}{3}}=(2)^\frac{2}{3}=(2^2)^\frac{1}{3}=4^{\frac{1}{3}}$