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A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+..........+\frac{2018}{2017^2+2017}\)
>\(\frac{2018}{2017^2+2017}+\frac{2018}{2017^2+2017}+........+\frac{2018}{2017^2+2017}\)
\(=\frac{2018}{2017^2+2017}.2017=\frac{2018.2017}{2017\left(2017+1\right)}=1\) (1)
Lại có:A<\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+1}+.........+\frac{2018}{2017^2+1}\)
\(=\frac{2018}{2017^2+1}.2017=\frac{2018.2017}{2017^2+1}=\frac{2017.\left(2017+1\right)}{2017^2+1}\)
\(=\frac{2017^2+2017}{2017^2+1}=\frac{2017^2+1+2016}{2017^2+1}=1+\frac{2016}{2017^2+1}< 2\) (2)
Từ (1) và (2) suy ra:1 < A < 2
Vậy A không phải là số nguyên
a) ta có: \(1-\frac{2016}{2017}=\frac{1}{2017}\)
\(1-\frac{2017}{2018}=\frac{1}{2018}\)
\(\Rightarrow\frac{1}{2017}>\frac{1}{2018}\Rightarrow1-\frac{2016}{2017}>1-\frac{2017}{2018}\Rightarrow\frac{2016}{2017}< \frac{2017}{2018}\)
b) ta có: \(\frac{2017}{2016}-1=\frac{1}{2016};\frac{2018}{2017}-1=\frac{1}{2017}\)
\(\Rightarrow\frac{1}{2016}>\frac{1}{2017}\Rightarrow\frac{2017}{2016}-1>\frac{2018}{2017}-1\Rightarrow\frac{2017}{2016}>\frac{2018}{2017}\)
\(=2017.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2017.\left(1-\frac{1}{100}\right)\)
\(=2017.\frac{99}{100}\)
\(=\frac{199693}{100}\)
\(\frac{2017}{1.2}+\frac{2017}{3.4}+\frac{2017}{4.5}+...+\frac{2017}{99.100}\) \(=2017.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\) \(=2017.\left(1-\frac{1}{100}\right)\) \(=2017.\frac{99}{100}\) \(=\frac{199693}{100}\)
