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 \(A=\dfrac{-4949}{19800}\)

Gọi A = 1.2.3 + 2.3.4 + ................... + 2011.2012.2013

4A = 1.2.3.4 + 2.3.4.4 + ...................... + 2011.2012.2013.4

4A = 1.2.3.4 + 2.3.4.(5 - 1) +..................... + 2011.2012.2013.(2014 - 2010)

4A = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .................... + 2011.2012.2013.2014 - 2010.2011.2012.2013

4A = 2011.2012.2013.2014

A = \(\frac{2011.2012.2013.2014}{4}=2011.503.2013.2014=..........\)

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

giải ra hết hộ mình nhé!

`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`

A = 1.2.3 + 2.3.4 + ... + 20.21.22

⇒ 4A = 1.2.3.4 + 2.3.4.4 + ... + 20.21.22.4

= 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) ... + 20.21.22.(23 - 19)

= 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + 3.4.5.6 + ... - 19.20.21.22 + 20.21.22.23

= 20.21.22.23

= 212520

⇒ A = 212520 : 4 = 53130

A=1x2x3 + 2x3x4 +…+ 100x101x102
Nhân A với 4 ta có :
A x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+100x101x102x4
A x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 100x101x102x(103 - 99)
A x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... + 100x101x102x103 - 99x100x1001x102
Sau khi cộng - trừ giản ước ta có : A x 4 = 100x101x102x103
A = 100 x101x102x103 : 4 = 26527650

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Bằng 26527650 đúng 100000000000000000000000000000%