Question

How do you find the sum of the geometric series `Sigma 81 (1/3)^(n-1)` n=1 to 7?

Answer 1

Desired sum is `121 4/9`

Explanation:

`Sigma81(1/3)^(n-1)` represents a geometric series whose first term is `a_1=81` and common ratio `r` is `1/3`. This can be easily found by putting values of `n` from `1` onward and series appears as

`81,27,9,3,1,....................`

The sum of such series up to first `n` terms is given by

`S_n=(81(1-r^n))/((1-r))`

Hence, in the given series the desired sum is

`(81(1-(1/3)^7))/((1-1/3))`

= `81(1-1/3^7)xx3/2`

= `81(2186/2187)xx3/2`

= `cancel81(cancel(2186)^1093/cancel(2187)^27)xx3/cancel2`

= `1093/9=121 4/9`