AMC 12B 2025 (Problem 7)What is the value of ∑n=2255log2(1+1n)(log2n)(log2(n+1))?\sum_{n=2}^{255}\frac{\log_2\left(1+\frac1n\right)}{(\log_2 n)(\log_2(n+1))}?n=2∑255(log2n)(log2(n+1))log2(1+n1)?(A) 34\text{(A)}\;\frac34(A)43(B) 1−1log2255\text{(B)}\;1-\frac{1}{\log_2 255}(B)1−log22551(C) 78\text{(C)}\;\frac78(C)87(D) 1516\text{(D)}\;\frac{15}{16}(D)1615(E) 1\text{(E)}\;1(E)1Related TopicsCoreToolkit 5 — Telescoping seriesToolkit 48 — Logarithms: Definition and PropertiesHints (5)Hint 1log2(1+1n)=log2(n+1n)=log2(n+1)−log2n.\log_2\left(1+\frac1n\right)=\log_2\left(\frac{n+1}{n}\right)=\log_2(n+1)-\log_2 n.log2(1+n1)=log2(nn+1)=log2(n+1)−log2n.Hint 2log2(1+1n)(log2n)(log2(n+1))\frac{\log_2\left(1+\frac1n\right)}{(\log_2 n)(\log_2(n+1))}(log2n)(log2(n+1))log2(1+n1)=log2(n+1)−log2n(log2n)(log2(n+1))=\frac{\log_2(n+1)-\log_2 n}{(\log_2 n)(\log_2(n+1))}=(log2n)(log2(n+1))log2(n+1)−log2n=1log2n−1log2(n+1).=\frac1{\log_2 n}-\frac1{\log_2(n+1)}.=log2n1−log2(n+1)1.Hint 3∑n=2255log2(1+1n)(log2n)(log2(n+1))\sum_{n=2}^{255}\frac{\log_2\left(1+\frac1n\right)}{(\log_2 n)(\log_2(n+1))}n=2∑255(log2n)(log2(n+1))log2(1+n1)=∑n=2255(1log2n−1log2(n+1))=\sum_{n=2}^{255}\left(\frac1{\log_2 n}-\frac1{\log_2(n+1)}\right)=n=2∑255(log2n1−log2(n+1)1)=1log22−1log23=\frac1{\log_2 2}-\frac1{\log_2 3}=log221−log231+1log23−1log24+\frac1{\log_2 3}-\frac1{\log_2 4}+log231−log241+1log24−1log25+\frac1{\log_2 4}-\frac1{\log_2 5}+log241−log251⋮\vdots⋮+1log2255−1log2256+\frac1{\log_2 255}-\frac1{\log_2 256}+log22551−log22561=1log22−1log2256.=\frac1{\log_2 2}-\frac1{\log_2 256}.=log221−log22561.Hint 4log22=1,log2256=log2(28)=8.\log_2 2=1,\qquad \log_2 256=\log_2(2^8)=8.log22=1,log2256=log2(28)=8.Hint 5By Hints 3 and 4, 11−18=78.\frac11-\frac18=\frac78.11−81=87.Final Answer(C) 78\frac7887