+595 Let
a + ar + ar^2 + ar^3 + \dotsb
be an infinite geometric series. The sum of the series is $2.$ The sum of the cubes of all the terms is $4.$ Find the common ratio.
+129845 a / (1 - r) = 2
a = 2(1 - r) → a^3 = 8 ( 1 - r)^3 (1)
And
a^3 + (ar)^3 + (ar^2)^3 + (ar^3)^3 + ...... = 4
a^3 / ( 1 - r^3) = 4
a^3 = 4 ( 1 - r^3) (2)
Equate (1), (2)
8 ( 1 - r)^3 = 4 (1 - r^3)
8 ( 1 - r)^3 = 4 ( 1 - r) (1 + r + r^2)
8 (1 -r)^2 = 4 ( 1 + r + r^2)
8 (r^2 - 2r + 1) = 4 ( 1 + r + t^2)
2r^2 - 4r + 2 = 1 + r + r^2
r^2 - 5r + 1 = 0
r = [ 5 - sqrt [ 25 - 4] ] / 2 = [ 5 - sqrt (21) ] / 2
