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2025 AMC 12B Problem 7

Problem 7 of 25EasierAlgebra

What is the value of ∑n=2255log⁡2(1+1n)(log⁡2n)(log⁡2(n+1))? \sum_{n=2}^{255} \frac{\log_2\left(1 + \frac{1}{n}\right)}{(\log_2 n)(\log_2(n+1))}?

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Solution

Let an=log⁡2n.a_n = \log_2 n. The numerator equals an+1−an,a_{n+1} - a_n, so each term is an+1−ananan+1=1an−1an+1.\dfrac{a_{n+1} - a_n}{a_n a_{n+1}} = \dfrac{1}{a_n} - \dfrac{1}{a_{n+1}}. Telescoping from n=2n = 2 to 255255 leaves 1log⁡22−1log⁡2256=1−18\dfrac{1}{\log_2 2} - \dfrac{1}{\log_2 256} = 1 - \dfrac{1}{8} =78.= \dfrac{7}{8}. Thus, the correct answer is C.
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Tagged: telescoping · logarithm

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