Answer
$\dfrac{63}{128}$
Work Step by Step
The general formula to find the nth partial sum of a Geometric sequence is given as: $ S_{n}=\dfrac{a(1-r^n)}{(1-r)}$
$ a_1=(\dfrac{1}{2})^{1+1}=\dfrac{1}{4}$ for $ i=1$
Now, $ S_{6}=\dfrac{(\dfrac{1}{4}) \times [1-(\dfrac{1}{2})^{6}]}{[1-(\dfrac{1}{2})]}=\dfrac{1}{4} \times (\dfrac{63}{32})$
Our answer is: $ S_{6}=\dfrac{63}{128}$
