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

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

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

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

Ta xét : \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)

\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)

...

\(\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)

Ta có : 2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

=> 2S = \(\frac{1}{1.2}-\frac{1}{99.100}\)

=> 2S = \(\frac{4949}{9900}\)

=> S = \(\frac{4949}{19800}\)

2S=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{98.99.100}\)

2S= \(1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)2S= 1- \(\dfrac{1}{100}\)

2S= \(\dfrac{99}{100}\)

S= \(\dfrac{99}{100}.\dfrac{1}{2}\)

S=\(\dfrac{198}{100}\)

1/2x3x4 + 1/3x4x5 + 1/4x5x6 + 1/5x6x7 + ..... + 1/8x9x10

= { 2/2x3x4 + 2/3x4x5 + 2/4x5x6 + .... + 2/8x9x10 } : 2

= { 4-2/2x3x4 + 5-3/3x4x5 + 6-4/4x5x6 + .... + 10-8/8x9x10 } : 2

= { 4/2x3x4 - 2/2x3x4 + 5/3x4x5 - 3/3x4x5 + ... + 10/8x9x10 - 8/8x9x10 } : 2

= { 1/2x3 - 1/3x4 + 1/3x4 - 1/4x5 + ... + 1/8x9 - 1/9x10 } : 2

=  { 1/2x3 - 1/9x10 } :2

=  { 1/6 - 1/90 } : 2

= 14/90 :  2

= 7/90

A = 1/2 - 1/3 - 1/4 + 1/3 - 1/4 - 1/5 + 1/4 - 1/5 - 1/6

A = 1/2 - 1/6

A = 1/3

k mk nha. ths bn nhìu nha

A = 1/2.3 - 1/ 3.4 + 1/3.4 - 1/4.5 + 1/4.5 -1/5.6

    = 1/2.3 - 1/5.6

    = 1/6 - 1/30

    =  2/15

Vậy A = 1/15

\(\frac{3x}{5}=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{6.7.8}\)

Ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{6.7.8}\)

      \(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{6.7.8}\right)\)

      \(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{6.7}-\frac{1}{7.8}\right)\)

      \(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{7.8}\right)\)

      \(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{56}\right)\)

      \(=\frac{1}{2}.\frac{27}{56}=\frac{27}{112}\)

\(\frac{3x}{5}=\frac{27}{112}\)

\(\Rightarrow3x=\frac{27.5}{112}\)

\(\Rightarrow3x=\frac{135}{112}\)

\(\Rightarrow x=\frac{45}{112}\)

~Học tốt~

Đặt A=\(\frac{-1}{2\cdot3\cdot4}\)+\(\frac{-1}{3\cdot4\cdot5}\)+...+\(\frac{-1}{12\cdot13\cdot14}\)

        A=(-1)*(\(\frac{1}{2\cdot3\cdot4}\)+\(\frac{1}{3\cdot4\cdot5}\)+...+\(\frac{1}{12\cdot13\cdot14}\))

        A=\(\frac{-1}{2}\)*(\(\frac{2}{2\cdot3\cdot4}\)+\(\frac{2}{3\cdot4\cdot5}\)+...+\(\frac{2}{12\cdot13\cdot14}\))

        A=\(\frac{-1}{2}\)*(\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)+\(\frac{1}{3\cdot4}\)-\(\frac{1}{4\cdot5}\)+...+\(\frac{1}{12\cdot13}\)-\(\frac{1}{13\cdot14}\))

       A=\(\frac{-1}{2}\)*(\(\frac{1}{6}\)-\(\frac{1}{182}\))

       A=\(\frac{-1}{2}\)*\(\frac{44}{273}\)

       A=\(\frac{-22}{273}\)

\(\frac{-1}{2.3.4}+\frac{-1}{3.4.5}+\frac{-1}{4.5.6}+...+\frac{-1}{11.12.13}+\frac{-1}{12.13.14}\)

\(=-\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+\frac{2}{4.5.6}+...+\frac{2}{11.12.13}+\frac{2}{12.13.14}\right)\)

\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{11.12}-\frac{1}{12.13}+\frac{1}{12.13}-\frac{1}{13.14}\right)\)

\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{13.14}\right)=-\frac{1}{2}.\frac{44}{273}=-\frac{22}{273}\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

\(2A=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

mình áp dụng công thức tổng quát:\(\frac{a}{n\left(n+1\right)\left(n+2\right)...\left(n+a\right)}=\frac{1}{n\left(n+1\right)\left(n+a-1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)...\left(n+a\right)}\)

Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

<=>\(2A=2\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\right)\)

<=>\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

<=>\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

<=>\(2A=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}=\frac{n^2+3n}{2\left(n+1\right)\left(n+2\right)}=\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}\)

<=>\(A=\frac{n\left(n+3\right)}{2\left(n+1\right)\left(n+2\right)}.\frac{1}{2}=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)

tổng quát:  1/n(n+1)(n+2)=1/2[1/n(n+1) - 1/(n+1)(n+2)]

ai k dung mik giai cho

\(D=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)

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

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

P/S:  tham khảo nhé

đến đây bn làm tiếp nha