Answered By Matt M

To solve this problem the first step it to recognise the pattern. The numerator increases by powers of two. Next, the series is rewritten to better represent the formula:

 $\sum_{k=0}^{n-1}ar^k=a\left(\frac{1-r^n}{1-r}\right)$∑_k=0ⁿ⁻¹ar^k=a(1−rⁿ1−r ) 

a is simply the first term: 2/3

r is the "common ratio" which we have already identified as: 2

n is the greatest power in the series:  $\sqrt{128}=7$√128=7 

and k is n-1: 6

Now, substituting these values for the variables, we get:

 $\sum_{k=0}^6\frac{2}{3}\left(2\right)^k=\frac{2}{3}\left(\frac{1-2^7}{1-2}\right)=\frac{2}{3}\left(\frac{-127}{-1}\right)=127\cdot\frac{2}{3}=84\frac{2}{3}=84.667$∑_k=0⁶23 (2)^k=23 (1−2⁷1−2 )=23 (−127−1 )=127·23 =8423 =84.667 

The sum of the geometric series is  $84\frac{2}{3}$8423  84.667