Question

What is the sum of the finite geometrice series?

3+9+27+81.....+6561

Answer 1

This is a geometric series that looks like;

`3+3(3)+3(3)^2+...+3(3)^7`
with
`a=3` and
`r=3`
And the sum of any geometric series =
`(a(1-r^(n+1) ))/(1-r)`
where n is the highest power, which is 7 in this case.
so,
The sum of this series=
`(3(1-3^8))/(1-3)= 9840`
Hope this helps!