Question

Calculate the following geometric sum. Round to three decimal places. (1.07)^18 + (1.07)^17 +...+ (1.07)^2 + (1.07) +1

Answer 1

Given:

`a=(1.07)^(18);\text{ r=}(1)/(1.07)`

The series power starts with 18 and ends in 0, n=19

`S_n=(a(1-r^n))/(1-r)`
`S_(19)=((1.07)^(18)(1-((1)/(1.07))^(19)))/(1-(1)/(1.07))`
`S_(19)=((1.07)^(18)\lbrack1-0.2766\rbrack)/((1.07-1)/(1.07))`
`S_(19)=((1.07)^(18)\lbrack0.7234\rbrack)/((0.07)/(1.07))`
`S_(19)=(1.07)^(18)\lbrack0.7234\rbrack*(1.07)/(0.07)`
`S_(19)=(1.07)^(19)\lbrack10.3343\rbrack`
`S_(19)=(3.6165)(10.3343)`
`S_(19)=37.3709`