cho \(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2017.2018}\) ; \(b=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}\) . Tính (a-b)^2019
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MT
9 tháng 4 2020
Đặt S = ( 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/2017.2018 )
Đặt A = ( 1/1.2 + 1/3.4 + ... + 1/2017.2018)
= 1 - 1/2 + 1/3 - 1/4 + ... + 1/2017 - 1/2018
= ( 1 + 1/3 + ... + 1/2017 ) - ( 1/2 + 1/4 + ... + 1/2018 )
= ( 1 + 1/2 + ... + 1/2018 ) - 2 ( 1/2 + 1/4 + ... + 1/2018) )
= ( 1 + 1/2 + ... + 1/2018 ) - ( 1 + 1/2 + ... + 1/1009 )
= 1/1010 + 1/1011 + ... + 1/2018
=> A - ( 1/1010 + 1/1011 + ... + 1/2017 ) = 1/2018
=> S = 1/2018
Vậy S = 1/2018
LM
13 tháng 8 2016
A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{2017.2018}\)
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2017}\)-\(\frac{1}{2018}\)
A=1-\(\frac{1}{2018}\)
A=\(\frac{2017}{2018}\)
13 tháng 8 2016
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2017.2018}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(A=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}+\frac{1}{2018}\)
Đến đây bình thường ta nhóm 2 số vào với nhau nhưng ở đây có lẻ số hạng nên không nhóm được => đề sai
29 tháng 6 2021
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
10 tháng 5 2018
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-1-\frac{1}{2}-...-\frac{1}{1009}\)
\(A=\frac{1}{1010}+\frac{1}{2000}+...+\frac{1}{2018}\)
\(B=3028.\left(\frac{1}{1010.2018}+...+\frac{1}{2018.1010}\right)\)
\(B=\frac{3028}{1010.2018}+...+\frac{3028}{2018.1010}\)
\(B=\frac{1}{1010}+\frac{1}{2018}+...+\frac{1}{2018}+\frac{1}{1010}\)
\(B=2.\left(\frac{1}{1010}+...+\frac{1}{2018}\right)\)
\(=>\frac{A}{B}=\frac{1}{2}\)
S
26 tháng 6 2019
\(\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)
\(theocaua\Rightarrow A=\frac{1}{26}+\frac{1}{27}+......+\frac{1}{50}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\left(5sohang\right)+\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\left(10sohang\right)+\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}\left(10sohang\right)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{35}{60}=\frac{7}{12}\left(1\right)\)
\(A=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}< \frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\left(5sohang\right)+\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}\left(10sohang\right)+\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\left(10sohang\right)=\frac{1}{4}+\frac{1}{3}+\frac{1}{5}=\frac{47}{60}< \frac{5}{6}=\frac{50}{60}\left(2\right)\) \(\left(1\right);\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)
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) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)
\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)
\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}.\frac{2013}{4030}\)
\(=\frac{6039}{8060}\)
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