Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)

\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)

hay \(A=\dfrac{-4949}{19800}\)

`1/(1.2.3) + 1/(2.3.4) +.....+ 1/(98.99.100)`

`2/(1.2.3) +  2/(2.3.4) + ...+ 2/(98.99.100)`

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

`1/(1.2) - 1/(99.100)`

`1/2 - 1/9900`

= `4949/9900`

=1/1x2-1/2x3+1/2x3-1/3x4+...+1/98x99-1/99x100

=1/2-1/9900

=4949/9900

Bằng 4949/9900

rõ ràng giùm cái

4A=1.2.3.4+2.3.4(5-1)+3.4.5(6-2)+.....+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+....+98.99.100.101-97.98.99.100

4A=98.99.100.101

A=(98.99.100.101):4=24497550

Cứ một dãy số  thì có 2 thừa số bị gạch nên cuối cùng chỉ còn 1x100

[ 98.99.100.101 - 0.1.2.3]: 4=24497550

Ta có:

\(F=1.2.3+2.3.4+...+98.99.100\)

\(\Rightarrow4F=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+....+98.99.100.\left(101-97\right)\)

\(\Rightarrow4F=1.2.3.4+2.3.4.5-1.2.3.4+...+98.99.100.101-97.98.99.100\)

\(\Rightarrow4F=98.99.100.101\Leftrightarrow F=\frac{98.99.100.101}{4}=24497550\)

(98x(98+1)x(98+2)x(98+3))=24497550

\(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{36\times37\times38}+\frac{1}{37\times38\times39}\)

\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{36\times37\times38}+\frac{2}{37\times38\times39}\)

\(2A=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)

\(2A=\frac{1}{1\times2}-\frac{1}{38\times39}\)

\(2A=\frac{741}{1482}-\frac{1}{1482}\)

\(2A=\frac{370}{741}\)

\(A=\frac{370}{741}:2=\frac{185}{741}\)

A=\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{37x38x39}\)

=\(\frac{1}{2}x\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{37x38}-\frac{1}{38x39}\right)=\frac{1}{2}x\left(\frac{1}{2}-\frac{1}{38x39}\right)=\frac{185}{741}\)

\(\frac{99}{100}\)nhé bạn ✿❀✿❀✿❀

Đặt A là tên biểu thức

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

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

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

\(2A=\frac{1}{2}-\frac{1}{9900}\)

\(2A=\frac{4949}{9900}\)

\(A=\frac{4949}{9900}:2=\frac{4949}{19800}\)

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

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

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

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

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

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