Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 8"
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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} | + | 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>. |
-xMidnightFirex | -xMidnightFirex |
Revision as of 15:21, 14 February 2020
Problem
For what integer value of is the expression equal to ? (Hint: )
Solution
Looking at the first fraction, we get that . Moving on, we see an interesting pattern in which the fraction . This means that we can rewrite the fractions as . We can cancel out most of the terms in that sequence and get . However, we need to solve for now. Setting the previous expression equal to , we get that . Squaring both sides, we get that . Hence, = .
-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 |