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TB
26 tháng 8 2018
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{20.21}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(=2\left(1-\frac{1}{21}\right)=2.\frac{20}{21}=\frac{40}{21}\)
KN
2
5 tháng 4 2018
\(A=\frac{1\cdot2+2\cdot3+3\cdot4+...+20\cdot21}{1+2-3-4+5+6-7-8+...+197+198-199-200+201}\) (1)
đặt \(B=1\cdot2+2\cdot3+3\cdot4+...+20\cdot21\)
\(3B=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+20\cdot21\cdot3\)
\(3B=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+20\cdot21\cdot\left(22-19\right)\)
\(3B=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+20\cdot21\cdot22-19\cdot20\cdot21\)
\(3B=20\cdot21\cdot22\)
\(B=\frac{20\cdot21\cdot22}{3}=3080\) (2)
đặt \(C=1+2-3-4+5+6-7-8+...+197+197-199-200+201\)
\(C=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(197+198-199-200\right)+201\)
\(C=-4+\left(-4\right)+...+\left(-4\right)+201\) có 50 số -4
\(C=-4\cdot50+201\)
\(C=-200+201\)
\(C=1\) (3)
\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow A=\frac{B}{C}=\frac{30801}{1}=3080\)
27 tháng 8 2017
a) = 1-1/2+1/2-1/3+1/3-1/4
= 1-1/4=3/4
b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018
=1-1/2018=2017/2018
c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015
= 1/2-1/2015=2015/4030-2/4030=2013/4030
Q
27 tháng 8 2017
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)
\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)
\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}.\frac{2013}{4030}\)
\(=\frac{6039}{8060}\)
NV
2
VA
19 tháng 7 2016
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{19.20}\)
\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=3.\left(1-\frac{1}{20}\right)\)
\(=3.\frac{19}{20}=\frac{57}{20}\)
Ủng hộ mk nha !!! ^_^
8 tháng 3 2017
\(S=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+....+\frac{3}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+......+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1}-\frac{1}{2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{2015}{2016}\)
\(\Rightarrow S=\frac{2015}{672}\)
Vậy: \(\Rightarrow S=\frac{2015}{672}\)
Bạn giải giúp mk câu mk đăng tầm 5 phút nha!
VH
5
AK
26 tháng 3 2018
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}\)
\(=\frac{297}{100}\)
Tham khảo nha !!!
6 tháng 4 2019
A=3(1/1.2+1/2.3+...+1/99.100)
A=3(1-1/2+1/2-1/3+...+1/99-1/100)
A=3(1-1/100)
A=3 . 99/100
A= 297 /100
5B= 1.2.3.4.5+2.3.4.5.5+....+97.98.99.100.5
=1.2.3.4.5+2.3.4.5.6 -1.2.3.4.5+...+-96.97.98.99
=97.98.99.100.101=9505049400
=> B=1901009880
BT
16 tháng 5 2017
Ta có: \(\frac{1}{1.2}=\frac{3}{1.2.3}\) ;\(\frac{1}{1.2+2.3}=\frac{3}{2.3.4}\); \(\frac{1}{2.3+3.4}=\frac{3}{3.4.5}\); ......;\(\frac{1}{1.2+2.3+3.4+...+n\left(n+1\right)}=\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(S=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Ta lại có: \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\); \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\); \(\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}\);....;\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)=> \(S=\frac{3}{4}-\frac{3}{2\left(n+1\right)\left(n+2\right)}< \frac{3}{4}\)
=> \(S< \frac{3}{4}\)
AK
8 tháng 4 2018
a ) \(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{201.203}\)
\(=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{201.203}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{201}-\frac{1}{203}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{203}\right)\)
\(=\frac{3}{2}.\left(\frac{203}{1015}-\frac{5}{1015}\right)\)
\(=\frac{3}{2}.\frac{198}{1015}\)
\(=\frac{297}{1015}\)
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Tham khảo nha !!!
Ta có : \(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{20.21}\)
\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{20.21}\right)\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(=3.\left(1-\frac{1}{21}\right)\)
\(=3.\frac{20}{21}\)
\(=\frac{20}{7}\)
Tham khảo nha !!! TK mk đi !!!
Ta có: \(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{20.21}\)
= \(3.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{20}-\frac{1}{21}\right)\)
= \(3.\left(\frac{1}{1}-\frac{1}{21}\right)\)
= \(3.\frac{20}{21}\)
= \(\frac{60}{21}\)