one over two plus one over four plus one over eight

one over two plus one over four plus one over eight

$$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} = ?\\\\
\boxed{
\dfrac{1}{2} = 2 * \left(\dfrac{1}{4} \right) \qquad
\dfrac{1}{4} = 2 * \left(\dfrac{1}{8} \right) }\\\\
\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} =
2*2* \left(\dfrac{1}{8}\right) + 2 * \left(\dfrac{1}{8}\right) +\left(\dfrac{1}{8}\right) = \left(\dfrac{1}{8}\right) \left(4+2+1\right) = \dfrac{7}{8}$$

What we have here is:

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{8}}}}$$

The easiest method to do this is to convert them all to even denominators.

We can do that by multiplying them by multiples of two.

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$ * $${\frac{{\mathtt{4}}}{{\mathtt{4}}}}$$ = $${\frac{{\mathtt{4}}}{{\mathtt{8}}}}$$

$${\frac{{\mathtt{1}}}{{\mathtt{4}}}}$$ * $${\frac{{\mathtt{2}}}{{\mathtt{2}}}}$$ = $${\frac{{\mathtt{2}}}{{\mathtt{8}}}}$$

Now that they all have a denominator of eight, we can esily add the numerators:

$${\frac{{\mathtt{4}}}{{\mathtt{8}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{8}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{8}}}}$$ = $${\frac{{\mathtt{7}}}{{\mathtt{8}}}}$$