Determine the sum of the even numbers between 1 and n

[lavekyl]

Determine the sum of the even numbers between 1 and n. n is specified by the user. Allow the user to repeat the process terminate if the user enters zero (0)
Example:
If the user enters 16, compute the sum of 2 + 4 + 6 + …16
If the user enters 21, compute the sum of 2 + 4 + 6 + …18 + 20
if the user enters 0 stop

Here is my code

#include <iostream>
using namespace std;


int main()
{  
    int n = 0;
    int j = 0;
    int sum = 0;
    char again = ' ';
    sum = 0;
    j = 2; 
    cout << "Enter a number: " ;
	cin >> n;
    while (j <= n)  
          {
                if (j%2==0)
                {
                   sum = sum + j;
                   j = j+2;         
                }
          }
          cout << "the sum of even numbers from 1 to " << n  << " is  " << sum << endl;
          cout << "Do you want to do this again (Y/N)?"; 
          cin >> again; 
          again = toupper(again);
                
    while (toupper(again) == 'Y')
          {
               ;
          }
                            
    system("pause");
	return 0;
}

This is as far as I have gotten. I have tried several things to make it work and can't seem to get it figured out.

[Xilonian]

you need a nested loop. Here is the psuedo code to get you going:

#include <iostream>
#include <iomanip>
using namespace std;


int main()
{  
    int n = 0;
    int j = 0;
    int sum = 0;
    char again = 'Y';
    sum = 0;
    j = 2; 
   
   
   	//  Loop 1: if again is y or Y then you want you go through this loop (remember or is designated ||)
	{ // opening bracket of first loop
		cout << "Enter a number: " ;
		cin >> n;
   		sum=0;
   		//Loop 2: Calculate sum of the even's. Start with j at 2 and increment it by 2 until it equals or is less than the number chosen by the user
		  
          cout << "the sum of even numbers from 1 to " << n  << " is  " << sum << endl;
          cout << "Do you want to do this again (Y/N)?"; 
          cin >> again;  //if they put in y or Y for this statement then the original loop will repeat
		  
	 }//closing bracket of first loop                     
   
	return 0;
}

[lavekyl]

I used what you said, and it didn't run the way it is supposed to.

[Xilonian]

This should do the trick

#include <iostream>
#include <iomanip>
using namespace std;


int main()
{  
	int n = 0;
	int j = 0;
	int sum = 0;
	char again = 'Y';
	sum = 0;
	j = 2; 
   
   
	while (again=='y' || again=='Y')
	{
		cout << "Enter a number: " ;
		cin >> n;
		sum=0;
	
		for(j=2;j<=n; j+=2)  
    	       {
    		sum += j;    	      
    	       }
		  
    	cout << "the sum of even numbers from 1 to " << n  << " is  " << sum << endl;
    	cout << "Do you want to do this again (Y/N)?"; 
    	cin >> again; 
		  
	}                     
	return 0;
}