Câu hỏi của Hung Tran

Question

Tính tổng sau:

A = $\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}$

help me!

Trần Minh Hoàng · Accepted Answer

$A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}$

$=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)$

$=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)$

$=2.\dfrac{502}{1005}=\dfrac{1004}{1005}$

Câu trả lời:

$A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}$

$=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)$

$=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)$

$=2.\dfrac{502}{1005}=\dfrac{1004}{1005}$

Ly Hoàng · Answer

$A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}$

$=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)$

$=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)$

$=2.\dfrac{502}{1005}=\dfrac{1004}{1005}$

Câu trả lời:

$A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}$

$=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)$

$=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)$

$=2.\dfrac{502}{1005}=\dfrac{1004}{1005}$