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business1
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
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?
great thanks tkhunny, i will study this!