1971 AMC 12 Problem 32

Question

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}

Set

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

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

1+x^{16}

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.

Problem 32 in other years