Question

How do you find the fifth partial sum of `sum_(i=1)^oo 2(1/3)^i`?

Answer 1

`sum_(i=1)^5 2(1/3)^i = 242/243`

Explanation:

This is a geometric series with initial term `a=2/3` and common ratio `r = 1/3`.

The general term of the series may be written:

`a_i = ar^(i-1)`

We find:

`(1-r) sum_(i=1)^n a_i = sum_(i=1)^n ar^(i-1) - r sum_(i=1)^n ar^(i-1)`

`color(white)((1-r) sum_(i=1)^n a_i) = sum_(i=1)^n ar^(i-1) - sum_(i=2)^(n+1) ar^(i-1)`

`color(white)((1-r) sum_(i=1)^n a_i) = a+color(red)(cancel(color(black)(sum_(i=2)^n ar^(i-1)))) - color(red)(cancel(color(black)(sum_(i=2)^n ar^(i-1)))) - ar^n`

`color(white)((1-r) sum_(i=1)^n a_i) = a(1-r^n)`

So dividing both ends by `(1-r)`, we find:

`sum_(i=1)^n a_i = (a(1-r^n))/(1-r)`

So in the given example, with `a=2/3` and `r=1/3`, we find:

`sum_(i=1)^5 2(1/3)^i = (2/3(1-(1/3)^5))/(1-(1/3))`

`color(white)(sum_(i=1)^5 2(1/3)^i) = (2/3(1-1/243))/(2/3)`

`color(white)(sum_(i=1)^5 2(1/3)^i) = 242/243`