Đặt C = \(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{30\times31\times32}\)

\(2C=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+...+\frac{2}{30\times31\times32}\)

        \(=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+...+\left(\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)

          \(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\)

            \(=\frac{1}{1\times2}-\frac{1}{31\times32}\)

              \(=\frac{1}{2}-\frac{1}{992}=\frac{495}{992}\)

\(\Rightarrow C=\frac{495}{992}\div2=\frac{495}{1984}\)

Vậy ...

\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+.....+\frac{1}{30\times31\times32}\)

\(=\frac{1}{2}\times\left(\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+.....+\frac{2}{30\times31\times32}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{30.31}-\frac{1}{31.32}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{31.32}\right)=\frac{1}{2}.\frac{990}{1984}=\frac{990}{3968}\)

\(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=190/380-1/380

=189/380

Gọi biểu thức trên là S. Ta có :

\(S=\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+\dfrac{1}{3\times4\times5}+...+\dfrac{1}{18\times19\times20}\)

\(=\dfrac{1}{2}\times\left(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{18\times19\times20}\right)\)

Trước tiên, ta áp dụng : \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)

Ta sẽ có : 

\(S=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{18\times19}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\dfrac{1}{1\times2}-\dfrac{1}{2}\times\dfrac{1}{19\times20}\)

\(=\dfrac{1}{4}-\dfrac{1}{760}=\dfrac{189}{760}\)

`=1/2(1/1×2 - 1/2×3 + 1/2×3 - 1/3×4 + 1/3×4 - 1/4×5 + ... + 1/18×19 - 1/19×20)`
`=1/2(1/2 - 1/19×20)`
`=1/2×189/380 `
`=189/760`

S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2013.2014.2015}\)

S = \(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{2015-2013}{2013.2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{2015}{2013.2014.2015}-\frac{2013}{2013.2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2014.2015}\right)\)

S = \(\frac{1}{2}.\frac{2029104}{4058210}\)

S = \(\frac{1014552}{4058210}\)

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

=1/2-1/9900

=4949/9900

Bằng 4949/9900

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Vũ Thị Trang lại là một nạn nhân tiếp tục bị báo cáo

`A=1/[1xx2xx3]+1/[2xx3xx4]+1/[3xx4xx5]+....+1/[98xx99xx100]`

`A=1/2xx(2/[1xx2xx3]+2/[2xx3xx4]+2/[3xx4xx5]+....+2/[98xx99xx100])`

`A=1/2xx(1/[1xx2]-1/[2xx3]+1/[2xx3]-1/[3xx4]+1/[3xx4]-1/[4xx5]+....+1/[98xx99]-1/[99xx100])`

`A=1/2xx(1/[1xx2]-1/[99xx100])`

`A=1/2xx(1/2-1/9900)`

`A=1/2xx(4950/9900-1/9900)`

`A=1/2xx4949/9900`

`A=4949/19800`

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

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

\(A=\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{12}-\dfrac{1}{20}+...+\dfrac{1}{9702}-\dfrac{1}{990}\right):2\)

\(A=\left(\dfrac{1}{2}-\dfrac{1}{990}\right):2\)

\(A=\dfrac{4949}{9900}:2\)

\(A=\dfrac{4949}{19800}\)

A = \(\dfrac{2}{1\times3\times5}\) + \(\dfrac{2}{3\times5\times7}\) + \(\dfrac{2}{5\times7\times9}\)+\(\dfrac{2}{7\times9\times11}\)

A = \(\dfrac{1}{2}\) x (\(\dfrac{4}{1\times3\times5}\) + \(\dfrac{4}{3\times5\times7}\) + \(\dfrac{4}{5\times7\times9}\) + \(\dfrac{4}{7\times9\times11}\))

A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\)-\(\dfrac{1}{3\times5}\)+\(\dfrac{1}{3\times5}\)-\(\dfrac{1}{5\times7}\)+\(\dfrac{1}{5\times7}\)-\(\dfrac{1}{7\times9}\)+\(\dfrac{1}{7\times9}\)-\(\dfrac{1}{9\times11}\))

A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\) - \(\dfrac{1}{9\times11}\))

A = \(\dfrac{1}{2}\) x (\(\dfrac{1}{3}-\dfrac{1}{99}\))

A = \(\dfrac{1}{2}\times\) \(\dfrac{32}{99}\)

A = \(\dfrac{16}{99}\)

B = \(\dfrac{1}{1\times2\times3}\) + \(\dfrac{1}{2\times3\times4}\) + \(\dfrac{1}{3\times4\times5}\) + \(\dfrac{1}{4\times5\times6}\)

B = \(\dfrac{1}{2}\) x (\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\dfrac{2}{4\times5\times6}\))

B = \(\dfrac{1}{2}\) x (\(\dfrac{1}{1\times2}\)-\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times3}\)-\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{3\times4}\)-\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{4\times5}\)-\(\dfrac{1}{5\times6}\))

B = \(\dfrac{1}{2}\)x(\(\dfrac{1}{1\times2}\) - \(\dfrac{1}{5\times6}\))

B = \(\dfrac{1}{2}\)x (\(\dfrac{1}{2}-\dfrac{1}{30}\))

B = \(\dfrac{1}{2}\)x \(\dfrac{7}{15}\)

B = \(\dfrac{7}{30}\)

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

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

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

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

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

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

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