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1977 AHSME Problems/Problem 24

Revision as of 15:09, 6 July 2016 by Dot22 (talk | contribs) (→‎Solution)

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

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