1977 AHSME Problems/Problem 24

Revision as of 15:07, 6 July 2016 by Dot22 (talk | contribs) (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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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 $1-\frac{1}{257}$.

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