1977 AMC 12 Problema 24

Question

Halla la suma

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

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}

Descompón

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

como diferencia de dos fracciones unitarias

Decompose

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

into two unit fractions

El último denominador corresponde a

n=128

The last denominator corresponds to

n=128

Para

1\le n\le128,

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

Por lo tanto, la suma telescópica es

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

Por lo tanto, la respuesta correcta es D.

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.

El Problema 24 en otros años