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, $\boxed{\textbf{(C)}\ 1 < x < 2}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 41
Followed by
Problem 43
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png