Difference between revisions of "1977 AHSME Problems/Problem 24"
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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> | + | 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 14:10, 6 July 2016
Solution
Note that 
. Indeed, we find the series telescopes and is equal to 
, which is evidently 
