编写4-8+12-16±n级数求和程序遇问题，请求排查代码错误

Let's break down the issues in your code and walk through how to fix them:

Key Problems in Your Current Code #

1. Incorrect Term Handling Logic
   Your loop calculates sum += i - (i+2) which assumes every pair of terms can be grouped together. But this fails when the series ends with a single positive term (e.g., when input n=4, the only term is 4, but your code computes 4-6=-2—way off the correct result of 4). This happens because i+2 isn't part of the series at all!

2. Misleading Output Description
   Your print statement says "The sum of %s first terms", but n is the last value in the series, not the number of terms. For example, if you input n=8, that's 2 terms, not 8—this will confuse anyone reading the output.

3. Variable Name Conflict
   Using sum as a variable name overwrites Python's built-in sum() function. While this doesn't break your code here, it's a bad practice that can cause issues later.

4. No Input Validation
   You don't check if n is actually a multiple of 4 (as required by the series rules). If someone inputs a number like 5, your code will still run and produce an incorrect result.

Corrected Solutions #

Solution 1: Iterate Through Each Term (Intuitive Approach) #

This method loops through each term in the series, applies the correct sign, and accumulates the sum:

n = int(input("Enter n: "))

# Validate input first
if n % 4 != 0:
    print("Error: n must be a multiple of 4!")
else:
    total = 0
    term_count = n // 4  # Calculate how many terms are in the series
    for k in range(1, term_count + 1):
        current_term = 4 * k
        # Apply positive sign to odd-numbered terms, negative to even
        if k % 2 == 1:
            total += current_term
        else:
            total -= current_term
    print(f"The sum of the series up to {n} is: {total}")

Solution 2: Use Mathematical Pattern (Efficient Approach) #

If you look closely at the series, you can spot a pattern to avoid looping entirely:

• When the number of terms (k = n//4) is even: The sum is -2 * k (e.g., 2 terms: 4-8=-4 = -22; 4 terms: 4-8+12-16=-8 = -24)
• When the number of terms is odd: The sum is 2 * (k + 1) (e.g., 1 term: 4 = 2*(1+1); 3 terms:4-8+12=8=2*(3+1))

Here's the code using this pattern:

n = int(input("Enter n: "))

if n % 4 != 0:
    print("Error: n must be a multiple of 4!")
else:
    k = n // 4
    if k % 2 == 0:
        total = -2 * k
    else:
        total = 2 * (k + 1)
    print(f"The sum of the series up to {n} is: {total}")

Both solutions will correctly calculate the sum for valid inputs. The first is great for understanding how the series builds, while the second is faster for very large values of n.

内容的提问来源于stack exchange，提问作者Anonymous