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7 tháng 5 2015

A = \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2014.2015.2016}\right)=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)=\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4062240}\right)=\frac{1}{4}-\frac{1}{8124480}

7 tháng 5 2015

Nhận xét: \(\frac{2}{1.2.3}=\frac{3-1}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)

\(\frac{2}{2.3.4}=\frac{4-2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)

........................

\(\frac{2}{2014.2015.2016}=\frac{2016-2014}{2014.2015.2016}=\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

=> \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2014.2015.2016}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\)

=> 2.A = \(2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2014.2015.2016}\right)=\frac{1}{1.2}-\frac{1}{2015.2016}

7 tháng 2 2020

H = \(\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)

   \(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)

Đặt G = \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

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

          = \(1-\frac{1}{100}\)

           = \(\frac{99}{100}\)

Đặt K = \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)

=>2K = \(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{99.100.101}\right)\)

          = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)

          = \(\frac{1}{1.2}-\frac{1}{100.101}\)

          = \(\frac{1}{2}-\frac{1}{10100}\)

          = \(\frac{5049}{10100}\)

=> K =\(\frac{5049}{10100}:2=\frac{5049}{10100}.\frac{1}{2}=\frac{5049}{20200}\)

Thay G,K vào H ta có :

H = \(\frac{99}{100}-\frac{5049}{20200}\)

Tự tính :)

7 tháng 2 2020

\(H=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.34}+...+\frac{1}{99.100.101}\right)\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\right)\)

\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\frac{99}{100}-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{99}{100}-\frac{1}{2}.\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)

11 tháng 3 2018

\(2C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{38.39}\)
\(C=\frac{617}{1482}\)

\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3D-D=1-\frac{1}{3^8}\)
\(D=\frac{1}{2}-\frac{1}{2.3^8}\)

11 tháng 3 2018

Ta có:\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{38.39}\right)\)

b,\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)

\(\Rightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)

\(\Rightarrow2D=1-\frac{1}{3^8}\)

\(\Rightarrow D=\frac{3^8-1}{3^8}:2\)

8 tháng 8 2016

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{90}\)

\(=\frac{22}{45}\)

2 tháng 4 2017

Gọi tổng trên là S , ta có :

S = 1/1.2.3 + 1/2.3.4 + ... + 1/8.9.10

S.2 = 2/1.2.3 + 1/2.3.4 + ... + 1/8.9.10

S.2 = 3 -1 /1.2.3 + 4 - 2/2.3.4 + ... + 10 - 8/8.9.10

S.2= 3/1.2.3 - 1/1.2.3 + 4/2.3.4 - 2/2.3.4 + ... + 10/8.9.10 - 8 /8.9.10

S.2 =1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/8.9 - 1/9.10

S.2 = 1/2 - 1/90

S = 1/4 - 1/360

S= 89/360

7 tháng 7 2017

\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\)

\(S=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\)

\(2S=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)

\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)

\(2S=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\)

\(\Rightarrow S=\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\div2=\frac{4949}{19800}\)

7 tháng 7 2017

Ta có:

Sx3 = 3/1 x ( 1/1x2x3 + 1/2x3x4 + .... + 1/98x99x100 )

Sx3 = 3/1x2x3 + 3/2x3x4 + .... + 3/98x99x100

Sx3 = (1/2 x 1/2x3) + (1/2x3 x 1/3x4) + ... + (1/98x99 + 1/99x100)

S     = (1/2 x 1/98x99) :3

S      = 1/59400

Mk ko quen vt p/s nên vt thế này cho nhanh sorry

24 tháng 4 2016

2A = 2/1.2.3 + 2/2.3.4 +.......+ 2/2014.2015.2016

2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 +.......+ 1/2014.2015 - 1/2015.2016

2A = 1/1.2 - 1/2015.2016

2A = ................Tự tính rồi tự làm nốt!

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 \(A=\dfrac{-4949}{19800}\)

8 tháng 8 2016

\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\)

  \(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)

     \(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{30}\right)=\frac{1}{4}.\frac{2}{15}=\frac{1}{30}\)

\(B=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{8.9}\)\(=5.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\)     \(=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)

  \(=5.\left(\frac{1}{2}-\frac{1}{9}\right)=5.\frac{7}{18}=\frac{35}{18}\)

\(C=\left(\frac{7^2}{2.9}+\frac{7^2}{9.16}+....+\frac{7^2}{65.72}\right):\left(\frac{1}{3}-\frac{7}{36}\right)\)

   \(=7.\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right):\frac{5}{36}\) \(=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right):\frac{5}{36}\)'

    \(=7.\left(\frac{1}{2}-\frac{1}{72}\right):\frac{5}{36}=7.\frac{35}{72}:\frac{5}{36}=\frac{49}{2}\)

\(D=\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}+\frac{2}{38.39.40}\)

     \(=2.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}+\frac{1}{38.39.40}\right)\)

     \(=2.\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}+\frac{1}{38.39}-\frac{1}{39.40}\right)\)

        \(=\frac{1}{2.3}-\frac{1}{39.40}=\frac{259}{1560}\)

\(E=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)

    \(=202202.\left(\frac{1}{3.4.101}+\frac{1}{4.5.101}+\frac{1}{5.6.101}+\frac{1}{6.7.101}+\frac{1}{7.8.101}\right)\)

      \(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)

        \(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)

         \(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)=2002.\frac{5}{24}=\frac{5005}{12}\)

     

    

3 tháng 8 2016

\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)

\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A< 1+1-\frac{1}{100}\)

\(A< 2-\frac{1}{100}< 2\)

4 tháng 8 2016

\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)

\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A< 1+1-\frac{1}{100}\)

\(A< 2-\frac{1}{100}< 2\)