Ask a New Question
Menu

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

1 answer

To print the sum of the first 10 numbers in the sequence \(3n - 1\), we want to make sure that for each \(n\) we correctly compute \(3n - 1\) and then sum that value from \(1\) to \(10\).

From the options provided, the correct altered code should look like this:

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

result = sum(10)
print(result)

Explanation:

  1. The base case is when \(n\) is 0, at which point we return 0 because there are no more numbers to sum.
  2. The recursive case computes the value of \(3n - 1\) for the current \(n\) and then calls itself with \(n - 1\) to continue summing up the previous values until it reaches the base case.
  3. We call sum(10) to compute the sum for the first 10 numbers of the sequence \(3n - 1\).

The other options do not correctly compute the sum for the desired sequence.