Root 2014?!

Let $q = \sqrt[2014]{2}$ . Then $\begin{aligned} x &= q^{2013} + q^{2012} + q^{2011} + \dots + q + 1 \\ &= \frac{q^{2014} - 1}{q - 1} \\ &= \frac{1}{\sqrt[2014]{2} - 1}. \end{aligned}$ We have therefore $\begin{aligned} \left( \frac{x+1}{x} \right)^{95} &= \left( 1 + \frac{1}{x} \right)^{95} \\ &= (1 + \sqrt[2014]{2} - 1)^{95} \\ &= \sqrt[2014]{2^{95}}. \end{aligned}$ And, finally, since $2014 = 19 \cdot 106$ and $95 = 19 \cdot 5$ , $\sqrt[2014]{2^{95}} = \sqrt[106]{2^{5}},$ and $a + b = \boxed{111}$ .

One should not simply borrow questions from SMO.

Anyway, multiply the equation by $\sqrt[2014]{2}$ .

Now, taking this and subtracting the original, we get $(\sqrt[2014]{2}-1)x=1$ . So now $1+\frac{1}{x}=\sqrt[2014]{2}$ and hence, we get our conclusion of 111