## Precalculus (10th Edition)

$\dfrac{1}{4}(2^n-1)$
We are given the sum: $\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{2^2}{4}+\dfrac{2^3}{4}+....+\dfrac{2^{n-1}}{4}$ Consider the geometric sequence: $a_1=\dfrac{1}{4}$ $r=2$ The given sum represents the sum $S_n$ of the first $n$ terms. Determine this sum: $S_n=a_1\dfrac{1-r^n}{1-r}=\dfrac{1}{4}\cdot\dfrac{1-2^n}{1-2}$ $=\dfrac{1}{4}(2^n-1)$