Write a function that computes future investment value at a given interest rate for a specified number of years.

Question

Use the following function header:

```
double futureInvestmentValue(
double investmentAmount, double monthlyInterestRate, int years)
```
For example, futureInvestmentValue(10000, 0.05/12, 5) returns 12833.59.

Write a test program that prompts the user to enter the investment amount (e.g., 1000) and the interest rate (e.g., 9%), and prints a table that displays future value for the years from 1 to 30, as shown below:

```

The amount invested: 1000
Annual interest rate: 9
Years Future Value
1 1093.8
2 1196.41
...
29 13467.25
30 14730.57

```

Answer 1

```
#include
#include
using namespace std;

double futureInvestmentValue(double investmentAmount, double monthlyInterestRate, int numOfYears)
{
return investmentAmount * pow(1 + monthlyInterestRate, numOfYears * 12);
}

int main()
{
cout << "Enter yearly interest rate, for example 8.25: ";
double annualInterestRate;
cin >> annualInterestRate;

cout << "Enter loan amount, for example 120000.95: ";
double loanAmount;
cin >> loanAmount;

cout << "Years\tFuture Value" << endl;

for (int i = 1; i <= 30; i++)
{
double futureValue = futureInvestmentValue(loanAmount, annualInterestRate / 1200, i);
cout << "" << i << "\t" << (int)(futureValue * 100) / 100.0 << endl;
}

return 0;
}
```