Difference between revisions of "1974 AHSME Problems/Problem 20"

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{{AHSME box|year=1974|num-b=19|num-a=21}}
 
{{AHSME box|year=1974|num-b=19|num-a=21}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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Latest revision as of 12:43, 5 July 2013

Problem

Let

\[T=\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}.\] (Error making remote request. Unexpected URL sent back)

Then

$\mathrm{(A)\ } T<1 \qquad \mathrm{(B) \ }T=1 \qquad \mathrm{(C) \  } 1<T<2 \qquad \mathrm{(D) \  } T>2 \qquad$

$\mathrm{(E) \  }T=\frac{1}{(3-\sqrt{8})(\sqrt{8}-\sqrt{7})(\sqrt{7}-\sqrt{6})(\sqrt{6}-\sqrt{5})(\sqrt{5}-2)}$

Solution

Let's try to rationalize $\frac{1}{\sqrt{n+1}-\sqrt{n}}$. Multiplying the numerator and denominator by $\sqrt{n+1}+\sqrt{n}$ gives us $\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}$.

Therefore,

\[T=(3+\sqrt{8})-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+2)=5.\]

Hence the answer is $\boxed{\text{D}}$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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
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