a) The balance of the account follows the next pattern:
[tex]\begin{gathered} f(0)=5000 \\ f(1)=5000(1+0.035)+5000=5000((1+0.035)+1) \\ f(2)=(5000(1+0.035)+5000)(1+0.035)+5000 \\ =5000(1+0.035)^2+5000(1+0.035)+5000 \\ =5000((1+0.035)^2+(1+0.035)+1) \\ \ldots \end{gathered}[/tex]
Where the function is f(t), t in quarters of a year.
10 years consists of 40 quarters; then, we need to calculate f(40),
[tex]\begin{gathered} f(40)=5000((1+0.035)^{40}+(1+0.035)^{39}+\cdots+(1+0.035)+1) \\ \Rightarrow f(40)=5000\sum ^{40}_{i\mathop=0}(1+0.035)^i \\ \Rightarrow f(40)\approx5000(88.51) \\ \Rightarrow f(40)\approx442550 \end{gathered}[/tex]
After 10 years, we will have approximately $442550
b) During those 10 years, we made 41 deposits of $5000 (taking into account the first deposit at t=0 and the final deposit at t=40)
[tex]41\cdot5000=205000[/tex]
The answer is $205000
c)
