1951 AHSME Problems/Problem 42

Problem

If $x =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$ , then:

$\textbf{(A)}\ x = 1\qquad\textbf{(B)}\ 0 < x < 1\qquad\textbf{(C)}\ 1 < x < 2\qquad\textbf{(D)}\ x\text{ is infinite}$ $\textbf{(E)}\ x > 2\text{ but finite}$

Solution

We note that $x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}=1+x$ . By the quadratic formula, $x=\dfrac{1\pm \sqrt{5}}{2}$ . Because there are only positive square roots in $x$ , $x$ must be positive, thus, it is $\dfrac{1+\sqrt{5}}{2}\approx 1.618$ . Thus, $\boxed{\textbf{(C)}\ 1 < x < 2}$ .

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 41	Followed by Problem 43
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1951 AHSME Problems/Problem 42

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