Problem 23 · 2018 Math Kangaroo
Stretch
Algebra & Patterns
arithmetic-sequencework-backward
The points \(A_0, A_1, A_2, \ldots\) all lie on a straight line. It is given that \(A_0A_1 = 1\) and that \(A_n\) is the midpoint of segment \(A_{n+1}A_{n+2}\) for every non-negative index n. How long is the segment \(A_0A_{11}\)?
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Answer: E — 683
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Hint 1 of 2
The midpoint rule rearranges to \(A_{n+2} = 2A_n - A_{n+1}\), so the signed step doubles and flips sign each time.
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Hint 2 of 2
Track the signed steps \(d_n = A_{n+1}-A_n\); they follow \(d_{n+1} = -2d_n\).
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Approach: find the recurrence for the directed steps, then sum
- Since \(A_n\) is the midpoint of \(A_{n+1}A_{n+2}\), the signed steps satisfy \(d_{n+1} = -2d_n\) with \(d_0 = 1\), so \(d_n = (-2)^n\).
- Then \(A_0A_{11} = |\,d_0 + d_1 + \cdots + d_{10}\,| = |1 - 2 + 4 - \cdots + 1024|\).
- This alternating geometric sum equals \(\frac{(-2)^{11}-1}{-3} = \frac{-2049}{-3} = 683\).
- So \(A_0A_{11} = \) 683.
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