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.

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