1951 AHSME Problems/Problem 42

Revision as of 08:37, 9 December 2011 by Yankeesfan (talk | contribs) (Created page with "== Problem == If <math> x =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}} </math>, then: <math> \textbf{(A)}\ x = 1\qquad\textbf{(B)}\ 0 < x < 1\qquad\textbf{(C)}\ 1 < x < 2\qquad\t...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 AHSME (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
All AHSME Problems and Solutions