Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 22"
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− | If you imagine expanding out the expression for <math>A</math>, you can see that every term in <math>C</math> will appear once, along with plenty of others. (Think of the prime | + | If you imagine expanding out the [[expression]] for <math>A</math>, you can see that every term in <math>C</math> will appear once, along with plenty of others. (Think of the [[prime factorization]]s and you can figure out which products give the terms of <math>C</math>.) Since all terms are [[positive]], <math>A > C</math>. |
− | <math> | + | <math>\frac AB = \left( 1 - \frac 12\right)\left( 1 + \frac 12 + \frac 14 + \frac 18 + \frac 1{16} \right) \left( 1 - \frac 13 \right)</math> <math>\left( 1 + \frac 13 + \frac 19\right)</math> <math>\left(1 - \frac 15\right) \left( 1 + \frac 15\right) \left(1 - \frac 17\right)</math> <math>\left( 1 + \frac 17\right) \left(1-\frac 1{11}\right)</math> <math>\left( 1 + \frac 1{11} \right)</math> <math>\left(1 - \frac 1{13}\right)</math> <math>\left( 1 + \frac 1{13}\right)</math> |
<math>= \left(1 - \frac1{32}\right)\left(1 - \frac1{27}\right)\left(1 - \frac1{25}\right) \left(1-\frac1{49}\right)\left(1-\frac1{121}\right)\left(1-\frac1{169}\right) < 1</math> so <math>A < B</math>. | <math>= \left(1 - \frac1{32}\right)\left(1 - \frac1{27}\right)\left(1 - \frac1{25}\right) \left(1-\frac1{49}\right)\left(1-\frac1{121}\right)\left(1-\frac1{169}\right) < 1</math> so <math>A < B</math>. |
Latest revision as of 01:42, 9 November 2007
Problem
Let
and
Then which of the following inequalities is true?
Solution
If you imagine expanding out the expression for , you can see that every term in will appear once, along with plenty of others. (Think of the prime factorizations and you can figure out which products give the terms of .) Since all terms are positive, .
so .
Putting these two facts together, .