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Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 8"

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== Solution ==
 
== Solution ==
  
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Looking at the first fraction, we get that <math>\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2} - \sqrt{1}</math>. Moving on, we see an interesting pattern in which the fraction <math>\frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}</math>. This means that we can rewrite the fractions as <math>(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{n+1} - \sqrt{n}</math>. We can cancel out most of the terms in that sequence and get <math>\sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1</math>. However, we need to solve for <math>n</math> now. Setting the previous expression equal to <math>7</math>, we get that <math>\sqrt{n+1} = 8</math>. Squaring both sides, we get that <math>n+1 = 64</math>. Hence, <math>n</math> = <math>\boxed{63}</math>.
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Looking at the first fraction, we get that <math>\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2} - \sqrt{1}</math>. Moving on, we see an interesting pattern in which the fraction <math>\frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}</math>. This means that we can rewrite the fractions as <math>(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{n+1} - \sqrt{n})</math>. We can cancel out most of the terms in that sequence and get <math>\sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1</math>. However, we need to solve for <math>n</math> now. Setting the previous expression equal to <math>7</math>, we get that <math>\sqrt{n+1} = 8</math>. Squaring both sides, we get that <math>n+1 = 64</math>. Hence, <math>n</math> = <math>\boxed{63}</math>.
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-xMidnightFirex
  
 
== See also ==
 
== See also ==

Latest revision as of 14:22, 14 February 2020

Problem

For what integer value of $n$ is the expression \[\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{n}+\sqrt{n+1}}\] equal to $7$ ? (Hint: $(1+\sqrt{2})(1-\sqrt{2})=-1.$)


Solution

Looking at the first fraction, we get that $\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2} - \sqrt{1}$. Moving on, we see an interesting pattern in which the fraction $\frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}$. This means that we can rewrite the fractions as $(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{n+1} - \sqrt{n})$. We can cancel out most of the terms in that sequence and get $\sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1$. However, we need to solve for $n$ now. Setting the previous expression equal to $7$, we get that $\sqrt{n+1} = 8$. Squaring both sides, we get that $n+1 = 64$. Hence, $n$ = $\boxed{63}$.

-xMidnightFirex

See also

1993 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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