Question

How do you find the sum of the first `13` terms of the geometric sequence: `7, 21, 63, 189, 567, . .`?

Answer 1

`5580127`

Explanation:

The general term of a geometric sequence can be written:

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

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

Then we find:

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

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

`= sum_(n=1)^N a r^(n-1) - sum_(n=2)^(N+1) a r^(n-1)`

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

`= a - a r^N`

`= a(1 - r^N)`

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

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

Note in particular that if `abs(r) < 1` then:

`lim_(N->oo) r^N = 0`

and:

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

Example

Given:

`7, 21, 63, 189, 567,...`

Note that:

`21/7 = 63/21 = 189/63 = 567/189 = 3`

That is, the sequence (as far as we are given) has a common ratio `r=3` between successive terms. The initial term is `a=7` and we are interested in the sum to `N=13` terms.

So the general term may be written:

`a_n = ar^(n-1) = 7 * 3^(n-1)`

and the sum to `13` terms is:

`sum_(n=1)^13 a_n = (a(1-r^N))/(1-r)`

`color(white)(sum_(n=1)^13 a_n) = (color(blue)(7) * (1-color(blue)(3)^color(blue)(13)))/(1-color(blue)(3)) = (7 * (1-1594323))/(1-3) = (-11160254)/(-2) = 5580127`