1971 AMC 12 Problema 32

Question

Si

\begin{aligned} s={}&(1+2^{-\frac{1}{32}})(1+2^{-\frac{1}{16}})\\ &\cdot(1+2^{-\frac{1}{8}})(1+2^{-\frac{1}{4}})\\ &\cdot(1+2^{-\frac{1}{2}}), \end{aligned}

entonces

s

es igual a:

If

\begin{aligned} s={}&(1+2^{-\frac{1}{32}})(1+2^{-\frac{1}{16}})\\ &\cdot(1+2^{-\frac{1}{8}})(1+2^{-\frac{1}{4}})\\ &\cdot(1+2^{-\frac{1}{2}}), \end{aligned}

then

s

is equal to:

\dfrac12(1-2^{-\frac{1}{32}})^{-1}

(1-2^{-\frac{1}{32}})^{-1}

1-2^{-\frac{1}{32}}

\dfrac12(1-2^{-\frac{1}{32}})

\frac{1}{2}

Sea

x=2^{-\frac{1}{32}}

Set

x=2^{-\frac{1}{32}}

Aplica repetidamente la identidad de diferencia de cuadrados hasta el factor

1+x^{16}

Apply the difference-of-squares identity repeatedly through the factor

1+x^{16}

Sea

x=2^{-\frac{1}{32}}.

Entonces

\begin{aligned} (1-x)s&=1-x^{32}\\ &=1-\frac12=\frac12. \end{aligned}

Por tanto,

s=\frac12(1-2^{-\frac{1}{32}})^{-1}.

Por lo tanto, la respuesta correcta es A.

Let

x=2^{-\frac{1}{32}}.

Then

\begin{aligned} (1-x)s&=1-x^{32}\\ &=1-\frac12=\frac12. \end{aligned}

Therefore

s=\frac12(1-2^{-\frac{1}{32}})^{-1}.

Therefore, the correct answer is A.

El Problema 32 en otros años