Question

How do you find the sum of the geometric series `7+21+63+...` to 10 terms?

Answer 1

`206668`

Explanation:

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

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

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

In our example, `a = 7` and `r = 3`.

Looking at the sum to `N` terms and multiplying by `(r-1)` we have:

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

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

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

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

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

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

In our example, we want the sum to `10` terms, so put `a=7`, `r=3` and `N=10` to find:

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

`color(white)(sum_(n=1)^N ar^(n-1))= (7(3^10 - 1))/(3-1) = (7*(59049 - 1))/2 = (7*59048)/2 = 206668`