Chứng minh: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\).
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Chứng minh :
\(\frac{1}{2}+\frac{1}{3.4}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Đặt tổng là S
\(\Rightarrow S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\)
\(\Rightarrow S=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(\Rightarrow S=\left(1+\frac{1}{2}+....+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(\Rightarrow S=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\) (đpcm)
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1. Chứng Minh Rằng \(\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+.....+\frac{100}{3^{100}}<\frac{3}{4}\)
2. Chứng Minh Rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2012}\)
2.
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\)
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Xem thêm câu trả lời
Chứng minh :
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
Mau nha,chiều nộp rồi.AI làm mau nhất và đứng nhất ,mình tick cho
chứng minh 1/1.2+1/3.4+1/5.6+...+1/2015.2016=1/1009+1/1010+...+1/2016
Chứng minh rằng:
a)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
b)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< 1-\frac{1}{2.3}\)
Cần gấp, ai nhanh mik tick nha
Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{2017.2018}\right)-\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+.....+\frac{1}{2017}\right)\)
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
thanks bạn nhiều
tính hợp lí:
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2017.2018}\right)-\left(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\right)\)
Chứng minh rằng : \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\)
Cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)Chứng minh \(\frac{7}{12}\)< A < \(\frac{5}{6}\)
Chứng minh \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)>\left(\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+...+\frac{1}{100}\right)\)
=> \(A>\frac{25}{75}+\frac{25}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)
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