Answer
$S_6=\dfrac{1365}{256} \approx 5.332$
Work Step by Step
Here, we have $\sum_{i=1}^6 4(1/4)^{i-1}$
We know that $S_{n}=a_1(\dfrac{1-r^{n}}{1-r})$
Now, $S_6=4[\dfrac{1-(1/4)^{6}}{1-(1/4)}]$
or, $=4 \times (\dfrac{4096/4096-1/4096}{3/4})$
or, $=4 \times \dfrac{1365}{1024}$
Hence, $S_6=\dfrac{1365}{256} \approx 5.332$