This doesn't look nice...

You guys are going to hate yourselves for how simple this is: upon close observation, it is possible to note that each coefficient of the sum is of the form $\left(- \frac{1}{2} \right)^i { -\frac{1}{3} \choose i }$ for integer values of $i \ge 1$ . This prompts us to consider the expansion of the binomial $(1+x)^{-\frac{1}{3}}$ , which is $1 - \frac{x}{3} + \frac{2 x^2}{9} - \frac{14x^3}{81} + \frac{35x^4}{243} - \ldots$

If we let $x = -\frac{1}{2}$ , the above expansion becomes

$1 + \sum _ { i = 1 } ^ { \infty } \frac {1 \times 4 \times 7 \times \ldots \times ( 3i - 2 ) } { 6 \times 12 \times 18 \times \ldots \times 6i } = (1-\frac{1}{2})^{-\frac{1}{3}} = \sqrt[3]{2} .$

It follows that our summation is equal to $\sqrt[3]{2} - 1 = \sqrt[15]{32} - \sqrt[15]{1}$ , so our answer is $\fbox{32}$

$\\$

A very similar problem can be found here .

Consider the following binomial expansion $2^{1/3}=(1-\frac{1}{2})^{-\frac{1}{3}}=1+\sum_{i=1}^{\infty}\frac{1}{i!2^i}\prod_{r=1}^{i}(\frac{1}{3}+r-1)\\=1+\sum_{i=1}^{\infty}\frac{1}{i!6^i}\prod_{r=1}^{i}(3r-2)$ Thus, $\sum_{i=1}^{\infty}\frac{1}{i!6^i}\prod_{r=1}^{i}(3r-2)=2^{1/3}-1=\sqrt[15]{32}-\sqrt[15]{1}$