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[nglinhrose]

Cho:

$A = \frac{1}{1.2} + \frac{1}{3.4} +.....+ \frac{1}{2003.2004} + \frac{1}{2005.2006}$

$B = \frac{1}{1004.2006} + \frac{1}{1005.2005} + \frac{1}{1006.2004} +.....+ \frac{1}{2006.1004}$

 

Hãy tính $\frac{A}{B}$

Edited by nglinhrose, 01-07-2016 - 22:07.

[hoakute]

$A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}=(1+\frac{1}{2}+...+\frac{1}{2006})-(1+\frac{1}{2}+..+\frac{1}{1003})=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}$

Lại có  $\frac{1}{3010}.B= \frac{1}{1004}+\frac{1}{2006}+\frac{1}{1005}+\frac{1}{2005}+...+\frac{1}{1004}=\frac{1}{1505}(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006})$

Suy ra A/B = 1505