44234
Explanation:[tex]\sum ^{97}_{n=1}80(1.03)^{n-1}[/tex]
The values of n is from 1 to 97.
We need to sum the result of each of the term from 1 to 97.
When n = 1
[tex]\begin{gathered} =80(1.03)^{1-1} \\ =80(1.03)^0\text{ = 80(1)} \\ =\text{ 80} \end{gathered}[/tex][tex]\begin{gathered} \text{when n = 2} \\ =80(1.03)^{2-1}\text{ = }=80(1.03)^1 \\ =\text{ 80(1.03) = 82.4} \\ \text{when n= 3} \\ =80\mleft(1.03\mright)^{\mleft\{3-1\mright\}}\text{ = }=80^{}\mleft(1.03\mright?^2) \\ =84.872 \end{gathered}[/tex]
Finding the sum:
[tex]\begin{gathered} \text{The sequence is a geoemtric sequence:} \\ \text{common ratio = next term/previous term = 1.03} \\ r\text{ >1} \\ \text{Sum of sequence in geometric sequence when r>1:} \\ S_n\text{ = }\frac{a(r^n-1)}{r-1} \end{gathered}[/tex][tex]\begin{gathered} n\text{ = 97} \\ r\text{ = }\frac{84.872}{84}=\frac{\text{ 84}}{80}\text{ = 1.03} \\ a\text{ = first term = 80} \\ S_{97}\text{ = }\frac{80(1.03^{97}-1)}{1.03-\text{ 1}} \\ \end{gathered}[/tex][tex]\begin{gathered} S_{97}\text{ = }\frac{80(16.5878)}{0.03} \\ S_{97}\text{ =}44234.1333 \end{gathered}[/tex]
To the nearest integer, the sum of 97 terms of the geometric sequence given is 44234.
