Difference between revisions of "1977 AHSME Problems/Problem 24"
(Created page with "==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>1-\frac{1}{257}</math>.") |
(→Solution) |
||
| (2 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
==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>1-\frac{1}{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 15:10, 6 July 2016
Solution
Note that 
. Indeed, we find the series telescopes and is equal to 
, which is evidently 
