Question

write a recursive racket function "sum-alternate" that takes a positive integer x as a parameter. The function should return the sum of all the integers x, x-2, x-4, x-6, etc. as long as the numbers are positive. For example, [sum-alternate 5] should evaluate to 5 + 3 + 1, and [sum-alternate 6] should evaluate to 6+4+2.

Answer 1

C++ Recursive Solution:

#include <iostream>
using namespace std;

//Recursive function 

int sumAlternate(int n, int i){
    
    //if we reach to zero simply return 2 as we have to reached to zero only if n
    // is at 2. and on substracting 2 it will become zero.
    if(n == 0) return 2;
    
    // if n is less than 0 return zero as we donot want to add this value
    if(n< 0) return 0;
    
    //or simply do recursion sum with substracting n by n-2;
     return n + sumAlternate(n-2*i, i+1);
     
}

//Main Method
int main() {
    
    int n;
    cout<<"Enter Number:";
    //take input value n
    cin>>n;
    
    //call recursive sumAlternate method to generate sum_alternate.
    //here 'i' have taken 1 as initial value so as to make formula
    //and 'i' will increase by 1 continously
    // x= x + x-(2*1) + x-(2*2) + x-(2*3) + ....... 
   cout<<"Alternate Sum is: "<<sumAlternate(n, 1);
        return 0;
}

Output:1.

output 2.