Maths Olympiad Prep

Track / Stage 4 / 70 of 340 #330 of 1964

Problem 330

AMC 12 late, AIME early
Algebra Difficulty 4.6 Prove it Bulgarian Winter Tournament · Bulgaria

Find A2024A_{2024}, where
An=1⋅2+3⋅4+5⋅8+⋯+(2n−1)⋅2n. A_n = 1 \cdot 2 + 3 \cdot 4 + 5 \cdot 8 + \dots + (2n-1) \cdot 2^n.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Since
2An=1⋅4+3⋅8+⋯+(2n−3)⋅2n+(2n−1)⋅2n+1, 2A_n = 1 \cdot 4 + 3 \cdot 8 + \dots + (2n-3) \cdot 2^n + (2n-1) \cdot 2^{n+1},
it follows
An=2An−An=(2n−1)⋅2n+1−(1⋅2+2⋅4+2⋅8+⋯+2⋅2n)=(2n−1)⋅2n+1−2⋅(2+4+8+⋯+2n)+1⋅2=(2n−1)⋅2n+1−2⋅2⋅2n−12−1+2=(2n−3)⋅2n+1+6. \begin{align*} A_n = 2A_n - A_n &= (2n-1) \cdot 2^{n+1} - (1 \cdot 2 + 2 \cdot 4 + 2 \cdot 8 + \dots + 2 \cdot 2^n) \\ &= (2n-1) \cdot 2^{n+1} - 2 \cdot (2 + 4 + 8 + \dots + 2^n) + 1 \cdot 2 \\ &= (2n-1) \cdot 2^{n+1} - 2 \cdot 2 \cdot \frac{2^n-1}{2-1} + 2 = (2n-3) \cdot 2^{n+1} + 6. \end{align*}
Therefore A2024=4045⋅22025+6A_{2024} = 4045 \cdot 2^{2025} + 6.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.