# Difference between revisions of "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 $\boxed \frac{128}{257}$ (Error compiling LaTeX. ! Argument of \frac has an extra }.)