The formula of the partial sum of the first n terms of a geometric series is:
[tex]Sn=\frac{a1(1-r^n)}{1-r}[/tex]
In this case, a1 is 6 (the first term of the series) and r is 1/4 (the base). By replacing 16 for n and the rest of the values we should get:
[tex]Sn=\frac{6(1-(\frac{1}{4})^{16})}{1-(\frac{1}{4})}\approx8[/tex]
Then, the sum of the given geometric series is 8
