Question

Use a while(true) loop to ask the user the following 2 values “Enter a rate r =” “Enter a nonnegative integer (enter negative integer to quit):”
If the user enters a negative int for n, the while loop is broken via the brake statement. Otherwise,
in the remaining part of the while loop, use a for loop to compute the partial sum for the geometric
series, namely 1 + r + rˆ2 + rˆ3 + . . . +rˆn.
Use the iomanip library in order to print the sum with 50 decimal digits of precision.

Sample output:

Enter a rate r = 4
Enter a non-negative integer (enter negative integer to quit): 9
1 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6 + 4^7 + 4^8 + 4^9 = 349525
Enter a rate r = 0
Enter a non-negative integer (enter negative integer to quit): 3
1 + 0^1 + 0^2 + 0^3 = 1
Enter a rate r = -4
Enter a non-negative integer (enter negative integer to quit): -2

C++

Will like if correct,thank you.

Answer 1

CODE -

#include<iostream>

#include<cmath>

#include<iomanip>

using namespace std;

int main()

{

    int r;

    int n;

    int sum;

    cout << setprecision(50);

    while(true)

    {

        // Take rate as input from the user

        cout << "Enter a rate r = ";

        cin >> r;

        // Take a non-negative integer as input from the user

        cout << "Enter a non-negative integer (enter negative integer to quit): ";

        cin >> n;

        // Exit the loop if user enters a negative integer

        if(n<0)

            break;

        // Initialize the sum to 1

        sum = 1;

        cout << sum;

        // Loop to compute partial sum for the geometric series and display the equation

        for(int i=1; i<=n; i++)

        {

            sum += pow(r, i);

            cout << " + " << r << "^" << i;

        }

        // Display the sum

        cout << " = " << sum << endl;

    }

}

SCREENSHOTS -

CODE -

OUTPUT -

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