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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{90}\)
\(=\frac{22}{45}\)
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
\(2C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{38.39}\)
\(C=\frac{617}{1482}\)
\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3D-D=1-\frac{1}{3^8}\)
\(D=\frac{1}{2}-\frac{1}{2.3^8}\)
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,\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
\(\Rightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)
\(\Rightarrow2D=1-\frac{1}{3^8}\)
\(\Rightarrow D=\frac{3^8-1}{3^8}:2\)
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 :)
\(H=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.34}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{99}{100}-\frac{1}{2}.\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
A = \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2014.2015.2016}\right)=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)=\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4062240}\right)=\frac{1}{4}-\frac{1}{8124480}<\frac{1}{4}\)
=> A < \(\frac{1}{4}\)
đúng cái
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 = \(2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2014.2015.2016}\right)=\frac{1}{1.2}-\frac{1}{2015.2016}<\frac{1}{2}\)
=> \(A<\frac{1}{4}\)
F = 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
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 \(A=\dfrac{-4949}{19800}\)
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\)
\(S=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\)
\(2S=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\)
\(\Rightarrow S=\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\div2=\frac{4949}{19800}\)
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