Difference between revisions of "1977 AHSME Problems/Problem 24"
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\textbf{(D) }\frac{128}{257}\qquad | \textbf{(D) }\frac{128}{257}\qquad | ||
\textbf{(E) }\frac{129}{257} </math> | \textbf{(E) }\frac{129}{257} </math> | ||
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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 <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> |
Latest revision as of 23:56, 1 January 2024
Find the sum .
Solution
Note that . Indeed, we find the series telescopes and is equal to
, which is evidently