Find the sum of the following series. Round to the nearest hundredth if necessary.

Answer:
11,184,808
Step-by-step explanation:
The n-th term of a geometric series is ...
an = a1·r^(n-1)
To fill in the formula, we need a1·r^n, so need to multiply the last term shown by r.
The value of r is 32/8 = 4, and the other terms of interest are a1 = 8, a1·r^(n-1) = 8388608. So, the sum is ...
[tex]S_n=\dfrac{ra_1r^{n-1}-a_1}{r-1}=\dfrac{4\cdot 8,388,608-8}{4-1}=11,184,808[/tex]