Given k, find the geometric sum i.e. 1 + 1/2 + 1/4 + 1/8 + … + 1/(2^k) using recursion.
Geometric Sum
PROBLEM:- Given k, find the geometric sum i.e. 1 + 1/2 + 1/4 + 1/8 + … + 1/(2^k) using recursion.
Input format :
Integer k
Output format :
Geometric sum
Constraints :
0 <= k <= 1000
Sample Input 1 :
3
Sample Output 1 :
1.875
Sample Input 2 :
4
Sample Output 2 :
1.93750SOLUTION:-
#include <iostream>
#include <math.h>
#include <iomanip>
using namespace std;
double geometricSum(int k) {
// Write your code here
if(k==0)
return 1;
else{
double sump= (1/(pow(2,k)));
return geometricSum(k-1)+sump;
}
}
int main() {
int k;
cin >> k;
cout << fixed << setprecision(5);
cout << geometricSum(k) << endl;
}