Question

How do you find `S_n` for the geometric series `a_1=2401`, r=-1/7, n=5?

Answer 1

`S_5 = 2101`

Explanation:

`color(white)()`
Background

The general term of a geometric series is described by the formula:

`a_n = ar^(n-1)`

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

Then we find:

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

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

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

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

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

`color(green)(sum_(n=1)^N a_n = (a(1 - r^N))/(1-r))`

`color(white)()`
Example

In our example, `a=2401=7^4`, `r = -1/7`, `N=5` and we find:

`sum_(n=1)^5 a_n = (color(blue)(2401)(1-(color(blue)(-1/7))^5))/(1-(color(blue)(-1/7)))`

`color(white)(sum_(n=1)^5 a_n) = (7^4(1+1/7^5))/(8/7)`

`color(white)(sum_(n=1)^5 a_n) = (7^4+1/7)/(8/7)`

`color(white)(sum_(n=1)^5 a_n) = (7^5+1)/8`

`color(white)(sum_(n=1)^5 a_n) = 16808/8`

`color(white)(sum_(n=1)^5 a_n) = 2101`

Alternatively, given the small number of terms, we could just add up the series:

`2401-343+49-7+1 = 2101`