TÍNH TỔNG:
\(S=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)
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Những câu hỏi liên quan
7 tháng 4 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
22 tháng 11 2015
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)x=\frac{1}{3}\left(2014.2015.2016-2013.2014.2015........+2.3.4-1.2.3+1.2.3-0.1.2\right)\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
\(x=\frac{1}{3.2029104}.2014^2.2015^2.2016=\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
7 tháng 3 2016
a)\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{100.101}\)=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+....+\(\frac{1}{100}\)-\(\frac{1}{101}\)=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)
b)\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+....+\(\frac{1}{28.29.30}\)=\(\frac{868}{3480}\)=\(\frac{217}{870}\)
c)\(\frac{1}{1.2.3.4}\)+\(\frac{1}{2.3.4.5}\)+....+\(\frac{1}{27.28.29.30}\)=\(\frac{24354}{438480}\)=\(\frac{451}{8120}\)
D
3
14 tháng 3 2017
=1+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{2}\) -\(\frac{1}{3}\) -\(\frac{1}{4}\)+\(\frac{1}{3}\) - \(\frac{1}{4}\)-\(\frac{1}{5}\)+.....+\(\frac{1}{99}\)-\(\frac{1}{100}\)-\(\frac{1}{101}\)
=1+\(\frac{1}{101}\)
=\(\frac{102}{101}\)
14 tháng 3 2017
1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]
1/2.3.4 = 1/2[ 1/2- 1/3 ]
...................
1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]
=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]
A = 1/2 . [1/1.2 -1/100 .101]
A= 1/2 . 5049 /10100 = 5049 / 20200.
Mình nghĩ là vậy đó.
TY
5 tháng 7 2017
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
DP
5 tháng 7 2017
a)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+\frac{5-3}{3\cdot4\cdot4}+....+\frac{100-98}{98\cdot99\cdot100}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)