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>.") |
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+ | Find the sum <math>\frac{1}{1(3)}+\frac{1}{3(5)}+\dots+\frac{1}{(2n-1)(2n+1)}+\dots+\frac{1}{255(257)}</math>. | ||
+ | |||
+ | <math>\textbf{(A) }\frac{127}{255}\qquad | ||
+ | \textbf{(B) }\frac{128}{255}\qquad | ||
+ | \textbf{(C) }\frac{1}{2}\qquad | ||
+ | \textbf{(D) }\frac{128}{257}\qquad | ||
+ | \textbf{(E) }\frac{129}{257} </math> | ||
+ | |||
==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> |
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