Question

find the sum of the infinite geometric series

1/3^6+1/3^8+1/3^10+1/3^12........

-100/9+10/3-1+3/10

Answer 1

a = 1/3^6
r = 1/3^2
S = a/(1-r) = (1/3^6)/(1-1/9) = (1/3^6)(9/8) = 1/648

a = -100/9
r = -3/10
S = (-100/9)/(1+3/10) = -1000/117

Answer 2

thank you