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Giải:
Ta có:
\(A=2\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\right).\)
\(A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}.\)
\(A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}.\)
\(A=\left(\dfrac{1}{2.3}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{3.4}-\dfrac{1}{3.4}\right)+...+\left(\dfrac{1}{98.99}-\dfrac{1}{98.99}\right)+\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right).\)
\(A=0+0+...+0+\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right).\)
\(A=\dfrac{1}{1.2}-\dfrac{1}{99.100}.\)
\(A=\dfrac{1}{2}-\dfrac{1}{9900}.\)
\(A=\dfrac{4950}{9900}-\dfrac{1}{9900}.\)
\(A=\dfrac{4949}{9900}.\)
Vậy \(A=\dfrac{4949}{9900}.\)
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Đặt A là tên biểu thức
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{9900}\)
\(2A=\frac{4949}{9900}\)
\(A=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Đặt A = 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 +....+ 98 x 99 x 100
4A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 4 + 4 x 5 x 4 +....+ 98 x 99 x 100 x 4
4A = 1 x 2 x 3 x ( 4 - 0 ) + 2 x 3 x 4 x ( 5 - 1 ) + 4 x 5 x 6 x ( 7 - 3 ) +....+ 98 x 99 x 100 x ( 101 - 97 )
4A = 1 x 2 x 3 x 4 + 2 x 3 x 4 x 5 - 1 x 2 x 3 x 4 + 4 x 5 x 6 x 7 - 3 x 4 x 5 x 6 + .... + 98 x 99 x 100 x 101 - 98 x 99 x 100 x 97
A = 98 x 99 x 100 x 97 / 4
A = 98 x 99 x 25 x 97
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+......+98.99.100.101-97.98.99.100
4a=98.99.100.101
a=(98.99.100.101):4=24497550
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
`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`
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 \(A=\dfrac{-4949}{19800}\)