Fraction Sum

Note that we can calculate it with usual process(by using sum of n terms of Geometric progression) but I like to provide pattern used here.

$\large\dfrac{1}{2} + \dfrac{1}{2} = 1$

$\large\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} = 1$

$\large\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{8} = 1$

$\large\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{16} = 1$

$\large\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} + \dfrac{1}{32} = 1$

$...$ So on

$\large\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} + \dfrac{1}{64} + \dfrac{1}{128} +\dfrac{1}{256} +\dfrac{1}{256} = 1$

Bonus question: Can you generalized this pattern?

Relevant wiki: Geometric Progression Sum

We are given that $\left(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} + \dfrac{1}{64} + \dfrac{1}{128} +\dfrac{1}{256}\right) +\dfrac{1}{x} = 1.$

Now using the formula for the sum of a geometric progression in the LHS of the equation, we get

$\left(\dfrac{\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^8\right]}{1-\frac{1}{2}}\right)+\dfrac{1}{x}=1$

$\Rightarrow \left(1-\left(\frac{1}{2}\right)^8\right)+\dfrac{1}{x}=1$

$\Rightarrow \dfrac{1}{x}=\dfrac{1}{2^8}$

$\therefore x=\boxed{256}$