AMC12 2025 B

AMC12 2025 B · Q7

AMC12 2025 B · Q7. It mainly tests Logarithms (rare), Algebra misc.

What is the value of \[\sum_{n = 2}^{255}\frac{\log_{2}\left(1 + \tfrac{1}{n}\right)}{\left(\log_{2}n\right)\left(\log_{2}(n + 1)\right)}?\]

求 \[\sum_{n = 2}^{255}\frac{\log_{2}\left(1 + \tfrac{1}{n}\right)}{\left(\log_{2}n\right)\left(\log_{2}(n + 1)\right)}\] 的值。

(A) \frac{3}{4} \frac{3}{4}

(B) 1 -\frac{1}{\log_{2}255} 1 -\frac{1}{\log_{2}255}

(D) \frac{15}{16} \frac{15}{16}

(E) 1 1

Answer

Correct choice: (C)

正确答案：（C）

Solution

By logarithm rules, write ∑n=2255log2⁡(1+1n)(log2⁡n)(log2⁡(n+1))=∑n=2255log2⁡(n+1n)(log2⁡n)(log2⁡(n+1))=∑n=2255log2⁡(n+1)−log2⁡n(log2⁡n)(log2⁡(n+1))=∑n=2255(1log2⁡n−1log2⁡(n+1)) But this telescopes to $\frac{1}{\log_{2}2}-\frac{1}{\log_{2}256}=\frac{1}{1}-\frac{1}{8}=\boxed{\textbf{(C)}~\frac{7}{8}}$.

由对数法则，写成 ∑_{n=2}^{255}\frac{\log_{2}(1+1/n)}{(\log_{2}n)(\log_{2}(n+1))}=∑_{n=2}^{255}\frac{\log_{2}((n+1)/n)}{(\log_{2}n)(\log_{2}(n+1))}=∑_{n=2}^{255}\frac{\log_{2}(n+1)-\log_{2}n}{(\log_{2}n)(\log_{2}(n+1))}=∑_{n=2}^{255}\left(\frac{1}{\log_{2}n}-\frac{1}{\log_{2}(n+1)}\right) 这是一个伸缩和，为 $\frac{1}{\log_{2}2}-\frac{1}{\log_{2}256}=\frac{1}{1}-\frac{1}{8}=\boxed{\textbf{(C)}~\frac{7}{8}}$。

Topics