Difference between revisions of "1977 AHSME Problems/Problem 24"

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==Solution==
 
==Solution==
  
Note that <math>\frac{1}{(2n-1)(2n+1)} = \frac{1/(2n-1)-1/(2n+1)}{2}</math>. Indeed, we find the series telescopes and is equal to <math>\frac{1-\frac{1}{257}}{2}</math>, which is evidently
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Note that <math>\frac{1}{(2n-1)(2n+1)} = \frac{1/(2n-1)-1/(2n+1)}{2}</math>. Indeed, we find the series telescopes and is equal to <math>\frac{1-\frac{1}{257}}{2}</math>, which is evidently <math>\boxed \frac{128}{257}</math>

Revision as of 15:09, 6 July 2016

Solution

Note that $\frac{1}{(2n-1)(2n+1)} = \frac{1/(2n-1)-1/(2n+1)}{2}$. Indeed, we find the series telescopes and is equal to $\frac{1-\frac{1}{257}}{2}$, which is evidently $\boxed \frac{128}{257}$ (Error compiling LaTeX. ! Argument of \frac has an extra }.)

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