Difference between revisions of "1959 AHSME Problems/Problem 37"
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<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> | ||
== 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 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>. |
== See also == | == See also == | ||
{{AHSME 50p box|year=1959|num-b=36|num-a=38}} | {{AHSME 50p box|year=1959|num-b=36|num-a=38}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:35, 21 July 2024
Problem 37
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 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
