Difference between revisions of "1959 AHSME Problems/Problem 37"
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| − | == Problem | + | == Problem == |
When simplified the product <math>\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)</math> becomes: | When simplified the product <math>\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)</math> becomes: | ||
| − | <math>\textbf{(A)}\ \frac1n \qquad\textbf{(B)}\ \frac2n\qquad\textbf{(C)}\ \frac{2(n-1)}{n}\qquad\textbf{(D)}\ \frac{2}{n(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)} </math> | + | <math>\textbf{(A)}\ \frac1n \qquad\textbf{(B)}\ \frac2n\qquad\textbf{(C)}\ \frac{2(n-1)}{n}\qquad\textbf{(D)}\ \frac{2}{n(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)} </math> |
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== Solution == | == Solution == | ||
The product <math>\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)</math> can be simplified to <math>\frac{2*3*4\cdots*(n-2)*(n-1)}{3*4*5\cdots*(n-1)*n}</math>, which, through [[telescoping series|telescoping]], ultimately works out to <math>\boxed{\textbf{(B) }\frac{2}{n}}</math>. | The product <math>\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)</math> can be simplified to <math>\frac{2*3*4\cdots*(n-2)*(n-1)}{3*4*5\cdots*(n-1)*n}</math>, which, through [[telescoping series|telescoping]], ultimately works out to <math>\boxed{\textbf{(B) }\frac{2}{n}}</math>. | ||
Latest revision as of 15:36, 21 July 2024
Problem
When simplified the product 
becomes:
Solution
The product can be simplified to 
, which, through telescoping, ultimately works out to 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 36 |
Followed by Problem 38 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
