Given k, find the geometric sum i.e.

PROBLEM:-

Geometric Sum

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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 :

1.93750

SOLUTION:-
#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;   
}

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