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:

$a=(1.07)^{18};\text{ r=}\frac{1}{1.07}$

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

$S_n=\frac{a(1-r^n)}{1-r}$$S_{19}=\frac{(1.07)^{18}(1-(\frac{1}{1.07})^{19})}{1-\frac{1}{1.07}}$$S_{19}=\frac{(1.07)^{18}\lbrack1-0.2766\rbrack}{\frac{1.07-1}{1.07}}$$S_{19}=\frac{(1.07)^{18}\lbrack0.7234\rbrack}{\frac{0.07}{1.07}}$$S_{19}=(1.07)^{18}\lbrack0.7234\rbrack\times\frac{1.07}{0.07}$$S_{19}=(1.07)^{19}\lbrack10.3343\rbrack$$S_{19}=(3.6165)(10.3343)$$S_{19}=37.3709$