1/1.2.3 + 1/2.3.4 +....+1/98.99.100

= 1/2 . (3-1/1.2.3 + 4-2/2.3.4 +....+ 100-98/98.99.100)

= 1/2 . (3/1.2.3 -1/1.2.3 + 4/2.3.4 - 2/2.3.4 +.......+ 100/98.99.100 - 98/98.99.100)

= 1/2 . (1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +......+ 1/98.99 - 1/99.100)

= 1/2 . (1/2 - 1/9900)

= 1/2 . 4949/9900

= 4949/19800

A=11.2.3+12.3.4+13.4.5+...+198.99.100=11.2−12.3+12.3−13.4+...+198.99−199.100=11.2−199.100=494919800

a/

\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)

\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)

b/

\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)

\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)

c/

\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)

vì bạn có l.i.k.e đâu mà trả lời

=1/1.2.3+1/2.3.4+1/3.4.5+............+1/98.99.100

 $=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)$

$=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)$

$=\frac{1}{2}\cdot \frac{4949}{9900}$

$=\frac{4949}{19800}$

cho mình xin lỗi vì đáp án mình gửi lên nó bị lỗi nhá

B=1/1.2.3+1/2.3.4+1/3.4.5+............+1/98.99.100

 \(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(=\frac{1}{2}\cdot\frac{4949}{9900}\)

\(=\frac{4949}{19800}\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(B=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

\(B=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(B=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)

\(\left(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\right)y=\dfrac{49}{200}\)

\(\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}+...+\dfrac{1}{98\cdot99}-\dfrac{1}{99\cdot100}\right)y=\dfrac{49}{200}\)

\(\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{99\cdot100}\right)y=\dfrac{49}{200}\)

\(\left(\dfrac{1}{4}-\dfrac{1}{19800}\right)y=\dfrac{49}{200}\)

\(\left(\dfrac{4950}{19800}-\dfrac{1}{19800}\right)y=\dfrac{49}{200}\)

\(\dfrac{4949}{19800}y=\dfrac{49}{200}\)

\(y=\dfrac{49}{200}:\dfrac{4949}{19800}\)

\(y=\dfrac{99}{101}\)

Vậy \(y=\dfrac{99}{101}\).

\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\right)y=\dfrac{49}{200}\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)y=\dfrac{49}{200}\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)y=\dfrac{49}{200}\\ \Rightarrow\dfrac{4949}{9900}y=\dfrac{49}{100}\\ \Rightarrow y=\dfrac{99}{101}\)

Chắc ko giòn đâu