What is the value of \sum_n=2^255 \frac\log_2(1 + (1)/(n))(\log_2 n)(\log_2(n+1))?

2025 AMC 12B Problem 7 — Solution

2025 AMC 12B Problem 7

Attempt Problem 7 of the 2025 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12B solutions, or check the answer key.

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What is the value of

$\sum_{n=2}^{255} \frac{\log_2\left(1 + \frac{1}{n}\right)}{(\log_2 n)(\log_2(n+1))}?$

$\dfrac{3}{4}$

$1 - \dfrac{1}{\log_2 255}$

$\dfrac{7}{8}$

$\dfrac{15}{16}$

$1$

Answer: C

Concepts:telescoping logarithm

Difficulty rating: 1420

Solution:

Let $a_n = \log_2 n.$ The numerator equals $a_{n+1} - a_n,$ so each term is $\dfrac{a_{n+1} - a_n}{a_n a_{n+1}} = \dfrac{1}{a_n} - \dfrac{1}{a_{n+1}}.$ Telescoping from $n = 2$ to $255$ leaves $\dfrac{1}{\log_2 2} - \dfrac{1}{\log_2 256} = 1 - \dfrac{1}{8}$ $= \dfrac{7}{8}.$

Thus, the correct answer is C.

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