\(=\dfrac{-1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{-1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{100}{101}=\dfrac{-50}{101}\)
`-2A = 2/(1.3) + 2/(3.5) + ... + 2/(99.101)`
`<=> -2A = 1/1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101`
`<=> -2A = 100/101`
`<=> x = -50/101`
`(-1)/(1.3) + (-1)/(3.5) + (-1)/(5.7) + ...... + (-1)/(99.101)`
Đặt `A = (-1)/(1.3) + (-1)/(3.5) + (-1)/(5.7) + ...... + (-1)/(99.101)`
`=> -2A = 2/(1.3) + 2/(3.5)+ 2/(5.7) + ...... + 2/99.101`
`<=> -2A = 1 -1/3 + 1/3 -1/4 + ....... + 1/99 - 1/101`
`-2A = 1 - 1/101`
`-2A = 101/101 - 1/101`
`-2A = 100/101`
`A= 100/101 :( -2)`
`A= 100/101 xx (-1/2)`
`A=(-50)/101`
`#LeMichael`