What is the value of ∑_n=2^255 log_2(1 + (1)/(n))(log_2 n)(log_2(n+1))?

2025 AMC 12B Problem 7

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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

Write

\log_2\!\left(1 + \frac{1}{n}\right)

= \log_2(n+1) - \log_2 n

With

a_n = \log_2 n,

each term becomes

\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}}

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

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.