Question

IN C++ AS SIMPLE AS POSSIBLE ______

Re-write the given function, printSeriesSquareFifth,  to use a while loop (instead of for).

• The function takes a single integer n as a parameter

• The function prints a series between 1 and that parameter, and also prints its result

• The result is calculated by summing the numbers between 1 and n (inclusive). If a number is divisible by 5, its square gets added to the result instead.

• The function does not return a value.

The series is: 1 + 2 + 3 + 4 + 5*5 +6 + 7 + 8 + 9 + 10*10 + 11 + 12 + 13 + 14 + 15*15 + 16 + ....

Example 1:
Calling the function with the integer 6 should print the following:
1 + 2 + 3 + 4 + 25 + 6
Result of the series is 41


Example 2:
Calling the function with the integer 15 should print the following:
1 + 2 + 3 + 4 + 25 + 6 + 7 + 8 + 9 + 100 + 11 + 12 + 13 + 14 + 225
Result of the series is 440

void printSeriesSquareFifth(int n) {
    int sum = 0;
    for (int i = 1;i <= n;i++) {
        if (i%5 == 0) {
            //If the number is divisible by 5, print its square and add the square to the sum
            cout << i*i;
            sum = sum + i*i;
        }
        else {
            //If the number is not divisible by 5, print the number and add the number to the sum
            cout << i ;
            sum = sum + i;
        }
        if(i != n){
            //This check is added to not print the last '+'
            cout<<" + ";
        }
    }
    cout << endl << "Result of the series is " << sum << endl;
}

Answer 1

void printSeriesSquareFifth(int n) {
    int sum = 0;
    int i = 1;
    while(i <= n) {
        if (i%5 == 0) {
            //If the number is divisible by 5, print its square and add the square to the sum
            cout << i*i;
            sum = sum + i*i;
        }
        else {
            //If the number is not divisible by 5, print the number and add the number to the sum
            cout << i ;
            sum = sum + i;
        }
        if(i != n){
            //This check is added to not print the last '+'
            cout<<" + ";
        }
        i++;
    }
    cout << endl << "Result of the series is " << sum << endl;
}