For this case we expand the series for each value of "n":
[tex]125 (\frac {1} {5}) ^ {1-1} +125 (\frac {1} {5}) ^ {2-1} +125 (\frac {1} {5}) ^ {3 -1} +125 (\frac {1} {5}) ^ {4-1} =[/tex]
[tex]125+ \frac {125} {5 ^ 1} + \frac {125} {5 ^ 2} + \frac {125} {5 ^ 3} =\\125+ \frac {125} {5} + \frac {125} {25} + \frac {125} {125} =\\125 + 25 + 5 + 1[/tex]
So, the correct option is: A
Answer:
Option A
