For this case, we must expand the series for each value of "n". That is to say:
![2 ^ {(1) -2} +2 ^ {(2) -2} +2 ^ {(3) -2} +2 ^ {(4) -2} +2 ^ {(5) -2} =](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mqcxnuwu2tkj44rfvg5qjk94qvowc2u3kz.png)
We have that equal signs are added and the same sign is placed, while different signs are subtracted and the sign of the major is placed:
2 ^ {1-2} + 2 ^ {2-2} + 2 ^ {3-2} + 2 ^ {4-2} + 2 ^ {5-2} =
![2 ^ {-1} + 2^(0)+ 2^(1) + 2^(2) + 2^(3)=](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o2m7mn29sch9zowukp256kk06wvy67oz3o.png)
By definition of power properties we have to:
![a ^ {-1} = \frac {1} {a}\\a ^ 0 = 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ci0jnn1qwjaider0os4x3u5hvyvyzty5if.png)
So:
![\frac {1} {2} + 1 + 2 + 4 + 8 =](https://img.qammunity.org/2020/formulas/mathematics/middle-school/yz0z6cym8cyd1rpmu31f586zz59byde1ps.png)
We find the common denominator:
![\frac {1} {2} + 1 * \frac {2} {2} + 2 * \frac {2} {2} + 4 * \frac {2} {2} + 8 * \frac {2} { 2} =\\\frac {1 + 2 + 4 + 8 + 16} {2} =\\\frac {31} {2} =\\15.5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jhuz89nm4pqaxa4ag7cqj2ua25u2rzebhr.png)
Answer:
Option B