1986 AMC 12 Problema 9

Question

El producto

\begin{aligned} &\left(1-\frac1{2^2}\right) \left(1-\frac1{3^2}\right)\\ &\quad{}\cdots \left(1-\frac1{9^2}\right) \left(1-\frac1{10^2}\right) \end{aligned}

es igual a

The product

\begin{aligned} &\left(1-\frac1{2^2}\right) \left(1-\frac1{3^2}\right)\\ &\quad{}\cdots \left(1-\frac1{9^2}\right) \left(1-\frac1{10^2}\right) \end{aligned}

equals

\frac5{12}

\frac12

\frac{11}{20}

\frac23

\frac7{10}

Factoriza

1-\frac{1}{n^2}

como una diferencia de cuadrados

Factor

1-\frac{1}{n^2}

as a difference of squares

Separa el producto en factores

\frac{n-1}{n}

y

\frac{n+1}{n}

Separate the product into factors

\frac{n-1}{n}

and

\frac{n+1}{n}

Para cada

n,

1-\frac1{n^2}=\frac{n-1}{n}\cdot\frac{n+1}{n}.

Por tanto, el producto se telescopa:

\begin{aligned} \prod_{n=2}^{10}\left(1-\frac1{n^2}\right) &=\left(\frac12\right) \left(\frac{11}{10}\right)\\ &=\frac{11}{20}. \end{aligned}

Por lo tanto, la respuesta correcta es C.

For each

n,

1-\frac1{n^2}=\frac{n-1}{n}\cdot\frac{n+1}{n}.

Therefore the product telescopes:

\begin{aligned} \prod_{n=2}^{10}\left(1-\frac1{n^2}\right) &=\left(\frac12\right) \left(\frac{11}{10}\right)\\ &=\frac{11}{20}. \end{aligned}

Thus the correct answer is C.

El Problema 9 en otros años