1977 AHSME Problems/Problem 24

Find the sum $\frac{1}{1(3)}+\frac{1}{3(5)}+\dots+\frac{1}{(2n-1)(2n+1)}+\dots+\frac{1}{255(257)}$ .

$\textbf{(A) }\frac{127}{255}\qquad \textbf{(B) }\frac{128}{255}\qquad \textbf{(C) }\frac{1}{2}\qquad \textbf{(D) }\frac{128}{257}\qquad \textbf{(E) }\frac{129}{257}$

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