编写Python程序计算等比数列1+2+4+…+n的和，参考指定代码格式

Got it, let's work through this problem. We need to compute the sum of a geometric series where each term doubles the previous one, starting at 1 and ending at the input value n.

First, a quick pattern check: each term here is a power of 2 (1=2⁰, 2=2¹, 4=2², etc.). The sum of this series up to 2ᵏ is actually 2ᵏ⁺¹ - 1—this is why when the input is 256 (which is 2⁸), the sum is 511 (2⁹ - 1). But let's build the solution following the iterative structure of the sample code you provided.

The Code #

n = int(input("Enter n: "))
total = 0
current_term = 1
# Loop through each term until we exceed the input n
while current_term <= n:
    total += current_term
    current_term *= 2  # Double the term to get the next in the sequence
print()
print("The sum is: %s" % (total))

How It Works #

• We start by grabbing the input n and initializing total (our sum accumulator) to 0.
• current_term starts at 1, the first term of the series.
• In each loop iteration, we add the current term to our total, then double the term to get the next value in the sequence.
• We stop looping once current_term grows larger than n, so we never add terms that exceed the input value.

Example Verification #

If you input 256, the loop processes terms 1, 2, 4, 8, 16, 32, 64, 128, 256. Adding these up gives 511, which matches the expected result.

Even if you input a value that's not a perfect power of 2 (like 10), the code works correctly: it sums 1+2+4+8 = 15, since 16 is larger than 10 and gets excluded from the sum.

内容的提问来源于stack exchange，提问作者 Anonymous