Find the sum &11×3+13×5+… & +1(2n-1)(2n+1)+… & +1255×257.

1977 AMC 12 Problem 24

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24.

Find the sum

\begin{aligned} &\frac1{1\cdot3}+\frac1{3\cdot5}+\cdots\\ &\quad+\frac1{(2n-1)(2n+1)}+\cdots\\ &\quad+\frac1{255\cdot257}. \end{aligned}

\frac{127}{255}

\frac{128}{255}

\frac12

\frac{128}{257}

\frac{129}{257}

Decompose

\frac1{(2n-1)(2n+1)}

into two unit fractions

The last denominator corresponds to

n=128

For

1\le n\le128,

\begin{gathered} \frac1{(2n-1)(2n+1)}\\ =\frac1{2(2n-1)}-\frac1{2(2n+1)}. \end{gathered}

Hence the sum telescopes to

\frac12\left(1-\frac1{257}\right)=\frac{128}{257}.

Therefore, the correct answer is D.