1977 AHSME Problems/Problem 24

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 $\boxed {\frac{128}{257}}$

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1977 AHSME Problems/Problem 24

Solution