LIVE by Po-Shen Loh标识

LIVE

中文

2025 AMC 12B Problem 7

Below is the professionally curated solution for Problem 7 of the 2025 AMC 12B, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:telescopinglogarithm

Difficulty rating: 1420

7.

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))}?

34\dfrac{3}{4}

1−1log⁡22551 - \dfrac{1}{\log_2 255}

78\dfrac{7}{8}

1516\dfrac{15}{16}

11

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=78.\dfrac{1}{\log_2 2} - \dfrac{1}{\log_2 256} = 1 - \dfrac{1}{8} = \dfrac{7}{8}.

Thus, the correct answer is C.

← Problem 6Full ExamProblem 8 →

Problem 7 in Other Years

1999 AMC 12 · 2000 AMC 12 · 2001 AMC 12 · 2002 AMC 12A · 2002 AMC 12B · 2003 AMC 12A · 2003 AMC 12B · 2004 AMC 12A · 2004 AMC 12B · 2005 AMC 12A · 2005 AMC 12B · 2006 AMC 12A · 2006 AMC 12B · 2007 AMC 12A · 2007 AMC 12B · 2008 AMC 12A · 2008 AMC 12B · 2009 AMC 12A · 2009 AMC 12B · 2010 AMC 12A · 2010 AMC 12B · 2011 AMC 12A · 2011 AMC 12B · 2012 AMC 12A · 2012 AMC 12B · 2013 AMC 12A · 2013 AMC 12B · 2014 AMC 12A · 2014 AMC 12B · 2015 AMC 12A · 2015 AMC 12B · 2016 AMC 12A · 2016 AMC 12B · 2017 AMC 12A · 2017 AMC 12B · 2018 AMC 12A · 2018 AMC 12B · 2019 AMC 12A · 2019 AMC 12B · 2020 AMC 12A · 2020 AMC 12B · 2021 AMC 12A Spring · 2021 AMC 12B Spring · 2021 AMC 12A Fall · 2021 AMC 12B Fall · 2022 AMC 12A · 2022 AMC 12B · 2023 AMC 12A · 2023 AMC 12B · 2024 AMC 12A · 2024 AMC 12B · 2025 AMC 12A