`sum_(i = 1)^12 16(1/2)^(i - 1)` Find the sum of the finite geometric series.

The given geometric series is:

`sum_(i=1)^12 16 (1/2)^(i-1)`

Since the summation notation is in the form

`sum_(i=1)^i a_1 *r^(i-1)`

it can be seen that the first term and common ratio of the geometric series is a1=16 and r=1/2.

Applying the summation formula of finite geometric series

`S_i=a_1* (1-r^i)/(1-r)`

the sum of the first twelve terms of this series is:

`S_12=16*(1-(1/2)^12)/(1-1/2)=4095/128`

Therefore,  `sum_(i=1)^12 16(1/2)^(i-1)=4095/128` .