Difference between revisions of "1986 AHSME Problems/Problem 9"
(Created page with "==Problem== The product <math> \left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\ldots\left(1-\frac{1}{9^{2}}\right)\left(1-\frac{1}{10^{2}}\right)</math> equals <ma...") |
(Added a solution with explanation) |
||
| Line 10: | Line 10: | ||
==Solution== | ==Solution== | ||
| − | + | Factor each term in the product as a difference of two squares, and group together all the terms that contain a <math>-</math> sign, and all those that contain a <math>+</math> sign. This gives <math>[(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{10})] \cdot [(1+\frac{1}{2})(1+\frac{1}{3})(1+\frac{1}{4})...(1+\frac{1}{10})] = [\frac{1}{2} \frac{2}{3} \frac{3}{4} ... \frac{9}{10}][\frac{3}{2} \frac{4}{3} \frac{5}{4} ... \frac{11}{10}] = \frac{1}{10} \cdot \frac{11}{2} = \frac{11}{20}</math>, which is <math>\boxed{C}</math>. | |
== See also == | == See also == | ||
Latest revision as of 17:27, 1 April 2018
Problem
The product 
equals
Solution
Factor each term in the product as a difference of two squares, and group together all the terms that contain a 
sign, and all those that contain a 
sign. This gives ![$[(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{10})] \cdot [(1+\frac{1}{2})(1+\frac{1}{3})(1+\frac{1}{4})...(1+\frac{1}{10})] = [\frac{1}{2} \frac{2}{3} \frac{3}{4} ... \frac{9}{10}][\frac{3}{2} \frac{4}{3} \frac{5}{4} ... \frac{11}{10}] = \frac{1}{10} \cdot \frac{11}{2} = \frac{11}{20}$](http://latex.artofproblemsolving.com/5/b/f/5bf29988be8d5119923d2ce558808d28dacc001f.png)
, which is 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
