Answer
2 - $\frac{1}{2^n}$
Work Step by Step
We know that,
$1 + r + r^2 + ··· + r^n = \frac{r^{n+1}− 1}{r − 1}$
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$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+ ... +\frac{1}{2^n}$
= $1+\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+ ... +(\frac{1}{2})^n$
= $\frac{(\frac{1}{2})^{n+1}− 1}{\frac{1}{2} − 1}$
=$\frac{(\frac{1}{2})^{n+1}− 1}{-\frac{1}{2}}$
=$2\times(1-(\frac{1}{2})^{n+1})$
=2 - $\frac{1}{2^n}$