/*
1.Write the sum of 1/(2^i) for i ranging from 0 to 2. 
2.Later generalise it to i ranging from 0 to N where N is a user defined input.
3.Challenge: Calculate 2^i without using pow function.
*/
#include <iostream>
#include <math.h>
using namespace std;
int main( )
{
 int         i =1 ;
 double      s= 0.0 ;// declaration

 for (i=0 ; i<= 2 ; ++i)
     {
       //s = s + user_function(s, i) ; // for task 3.
         s = s + pow (0.5,i); // ( 0.0 + 0.5^0)  i=0, s=1
      cout << "sum = " << s << endl;
     }//for- loop



return 0;

}