Recursive Algorithms Quick Check 2 of 52 of 5 Items Question Given the code for the recursive algorithm that sums the sequence of numbers from one to one hundred: def sum(n): if n < 1: return 0 else: return n + sum(n - 1) result = sum(100) print(result) How could the code be altered to print the sum of the first 10 numbers in the sequence 3n−1 ? (1 point) Responses def sum(n): if n <= 0: return 0 else: return n + sum(3 * n - 1) result = sum(10) print(result) def sum(n): if n <= 0: return 0 else: return n + sum(3 * n - 1) result = sum(10) print(result) def sum(n): if n == 0: return 0 else: return 3 * n - 1 + sum(n - 1) result = sum(100) print(result) def sum(n): if n == 0: return 0 else: return 3 * n - 1 + sum(n - 1) result = sum(100) print(result) def sum(n): if n == 0: return 0 else: return 3 * (n - 1) + sum(n - 1) result = sum(10) print(result) def sum(n): if n == 0: return 0 else: return 3 * (n - 1) + sum(n - 1) result = sum(10) print(result) def sum(n): if n == 0: return 0 else: return 3 * n - 1 + sum(n - 1) result = sum(10) print(result) def sum(n): if n == 0: return 0 else: return 3 * n - 1 + sum(n - 1) result = sum(10) print(result) Skip to navigation page 9 of 9

Answer 1

To print the sum of the first 10 numbers in the sequence defined by the expression \(3n - 1\), the correct implementation of the recursive algorithm should correctly compute the sum using the specified formula for each integer from 1 to 10.

The correct code needs to:

1. Start from \(n = 1\) to \(n = 10\).
2. Use the formula \(3n - 1\) for summation.
3. Recursively call the function to sum until \(n\) reaches zero.

Let's analyze the provided options:

1. def sum(n): 
       if n <= 0: 
           return 0 
       else: 
           return n + sum(3 * n - 1) 
   result = sum(10) 
   print(result)
   

   • This is incorrect because it incorrectly modifies the recursion.

2. def sum(n): 
       if n <= 0: 
           return 0 
       else: 
           return n + sum(3 * n - 1) 
   result = sum(10) 
   print(result)
   

   • Same as 1. Incorrect.

3. def sum(n): 
       if n == 0: 
           return 0 
       else: 
           return 3 * n - 1 + sum(n - 1) 
   result = sum(100) 
   print(result)
   

   • This is incorrect as it sums numbers up to 100 rather than 10 and uses the correct formula but does not fit the requirements.

4. def sum(n): 
       if n == 0: 
           return 0 
       else: 
           return 3 * n - 1 + sum(n - 1) 
   result = sum(100) 
   print(result)
   

   • This is still incorrect similar to the previous one.

5. def sum(n): 
       if n == 0: 
           return 0 
       else: 
           return 3 * (n - 1) + sum(n - 1) 
   result = sum(10) 
   print(result)
   

   • This incorrectly computes the terms since it uses \(3(n-1)\), which does not match the intended sequence.

6. def sum(n): 
       if n == 0: 
           return 0 
       else: 
           return 3 * (n - 1) + sum(n - 1) 
   result = sum(10) 
   print(result)
   

   • Similar to 5, this is incorrect for the same reasons.

7. def sum(n): 
       if n == 0: 
           return 0 
       else: 
           return 3 * n - 1 + sum(n - 1) 
   result = sum(10) 
   print(result)
   

   • This is correct! It computes \(3n - 1\) correctly and sums up the values from 1 through 10.

Correct answer:

def sum(n): 
    if n == 0: 
        return 0 
    else: 
        return 3 * n - 1 + sum(n - 1) 
result = sum(10) 
print(result)

This code will correctly print the sum of the first 10 numbers in the sequence \(3n - 1\).