Question

What is the sum of an infinite geometric series with a first term of 6 and a common ratio of 1/2?

Answer 1

`12`

Explanation:

The general term of a geometric series can be represented by the formula:

`a_n = a r^(n-1)`

where `a` is the initial term and `r` the common ratio.

We find:

`(1-r) sum_(n=1)^N ar^(n-1) = sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)`

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

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

`color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a(1 - r^N)`

Dividing both ends by `(1-r)` we find:

`sum_(n=1)^N ar^(n-1) = (a(1 - r^N))/(1-r)`

If `abs(r) < 1` then `lim_(N->oo) r^N = 0` and we find:

`sum_(n=1)^oo ar^(n-1) = lim_(N->oo) (a(1 - r^N))/(1-r) = a/(1-r)`

In the given example, `a=6` and `r=1/2`. So `abs(r) < 1` and:

`sum_(n=1)^oo 6*(1/2)^(n-1) = 6/(1-1/2) = 12`