Answer
$\dfrac{1}{4}(2^n-1)$
Work Step by Step
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:
$S_n=a_1\dfrac{1-r^n}{1-r}=\dfrac{1}{4}\cdot\dfrac{1-2^n}{1-2}$
$=\dfrac{1}{4}(2^n-1)$
