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Difference between revisions of "1977 AHSME Problems/Problem 24"
 
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\textbf{(D) }\frac{128}{257}\qquad
 
\textbf{(D) }\frac{128}{257}\qquad
 
\textbf{(E) }\frac{129}{257} </math>  
 
\textbf{(E) }\frac{129}{257} </math>  
−
 
  
 
==Solution==
 
==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>\frac{1-\frac{1}{257}}{2}</math>, which is evidently <math>\boxed {\frac{128}{257}}</math>
 
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>\frac{1-\frac{1}{257}}{2}</math>, which is evidently <math>\boxed {\frac{128}{257}}</math>

Latest revision as of 22:56, 1 January 2024

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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