Question

Construct a recursive definition for f(n) = floor(0/2) + floor(1/2) + … + floor(n/2), where the variables represent natural numbers.

Answer 1

C++ code:

#include<iostream>
#include<math.h>
using namespace std;

int func(double n){

if (n == 0){
return floor(n/2);
}
else{
return floor(n/2) + func(n-1);
}
}

int main(){
int n=10;
cout<<"f(n) for n = "<<n<<" is "<< func(n)<<"\n";

}

Screenshot of above code:

1 2 3 4 #include<iostream> #include<math.h> using namespace std; lint func (double n) { 7 8 9 10 11 if (n == 0) { return floor (n/2); } else{ return floor (n/2) + func(n-1); } 12 13 -} 14 15 16 int main() { int n=10; cout<<"f(n) for n = "<<n<<" is "<< func(n) <<"\n"; 17 18 19 } 20

Output:

f(n) for n = 10 is 25