S = 2*4+4*6+6*8+...+46*48+48*50

S6 = 2*4*6+4*6*6+6*8*6+........................+46*48*6+48*50*6

S6=2*4*(6-0)+4*6*(8-2)+6*8*(10-4)+.................................+46*48*(50-44)+48*50*(52-46)

S6 = 2*4*6+4*6*8-2*4*6+6*8*10-4*6*8+..........................................+46*48*50-44*46*48+48*50*52-46*48*50

S6 = 48*50*52=124800

S=124800/6=20800

\(S=2\cdot4+4\cdot6+...+48\cdot50\)

\(S=2\left(1\cdot2+2\cdot3+...+24\cdot25\right)\)

\(\Rightarrow3S=2\left(1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+...+24\cdot25\left(26-23\right)\right)\)

\(\Rightarrow3S=2\left(1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+...+24\cdot25\cdot26-23\cdot24\cdot25\right)\)

\(\Rightarrow3S=2\cdot24\cdot25\cdot26\)

\(\Rightarrow S=2\cdot8\cdot25\cdot26=10400\)

\(\Rightarrow6S=10400\cdot6=62400\)

bậy bạ

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\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=1-\dfrac{1}{25}\)

\(=\dfrac{24}{25}\)

Nhanh giúp mình với cả nhà ơi

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\dfrac{49}{50}< 1\)

Đáp án =2525 vì câu của cậu có người hỏi rồi

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{2.2004}{2010}=\frac{2004}{1005}\)

\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{1004\cdot1005}\)

\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1004\cdot1005}\right)\)

\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)

\(=2\cdot\left(1-\frac{1}{1005}\right)=2\cdot\frac{1004}{1005}=\frac{2008}{1005}\)

\(B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{98.100}\)

\(\Rightarrow5B=\dfrac{20}{2.4}+\dfrac{20}{4.6}+...+\dfrac{20}{98.100}\)

\(\Rightarrow5B=10\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}\right)\)

\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(\Rightarrow5B=10.\dfrac{49}{100}\)

\(\Rightarrow5B=\dfrac{49}{10}\)

Vậy \(5B=\dfrac{49}{10}\)

Ta có: B = \(\dfrac{4}{2.4}\) + \(\dfrac{4}{4.6}\) + \(\dfrac{4}{6.8}\) + ... + \(\dfrac{4}{98.100}\).

=> \(\dfrac{B}{2}\) = \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) + \(\dfrac{2}{6.8}\) + ... + \(\dfrac{2}{98.100}\)

=\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\)

= \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\) = \(\dfrac{49}{100}\).

=> B = \(\dfrac{49}{200}\).

=> 5B = \(\dfrac{49}{200}\) . 5 = \(\dfrac{49}{40}\).

Vậy 5B = \(\dfrac{49}{40}\).