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>\textbf{(B)} \frac{2}{n}</math>
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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>
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== 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 15:33, 21 July 2024

Problem 37

When simplified the product $\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)$ becomes: $\textbf{(A)}\ \frac1n \qquad\textbf{(B)}\ \frac2n\qquad\textbf{(C)}\ \frac{2(n-1)}{n}\qquad\textbf{(D)}\ \frac{2}{n(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)}$

Solution

The product $\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)$ can be simplified to $\frac{2*3*4\cdots*(n-2)*(n-1)}{3*4*5\cdots*(n-1)*n}$, which ultimately works out to $\boxed{\textbf{(B) }\frac{2}{n}}$

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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