Question

how to sum geometric series 5*(1,06)^-3 + 5*(1,06)^-4 +...+ 5*(1,06)^n-1, so as to end up with 5*((1-(1,06)^1-n) / (1,06^2 - 1,06)). thanks

Answer 1

Hint - It's an old method with some application.

1) \(r + r^{2} + r^{3} + ... + r^{n} = S\)

Multiply by r

2) \(r^{2} + r^{3} + r^{4} + ... + r^{n+1} = Sr\)

Subtract 2) from 1)

\(r - r^{n+1} = S - Sr\)

Can you solve for S?

Answer 2

great thanks tkhunny, i will study this!