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B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)
B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022
B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\)
B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\)
B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))
Vậy B > C
Đặt dãy trên có tổng là A
Ta có A= (0/1+2)(0/1+2+3). ... .(0/1+2+3+...+2022)
=> A=0+0+0+...+0+0+0=0
do 1/2= 0,5 và 0,5 >0 => 0>A
\(2022A=2022+2022^2+2022^3+2022^4+...+2022^{2018}\)
\(2021A=2022A-A=2022^{2018}-1\Rightarrow A=\dfrac{2022^{2018}-1}{2021}\)
\(\Rightarrow A< B\)
\(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2022^2}\)
\(C=\dfrac{1}{\left(2+3+4+...+2022\right)^2}\)
\(C=\dfrac{1}{\left(2022-2+1\right)^2}\)
\(C=\dfrac{1}{2021^2}\)
\(C=\dfrac{1}{2021\cdot2021}\)
\(C=\dfrac{1\div2021}{2021}\)
\(C=1\)
Vì \(1>\dfrac{13}{18}\)
\(\Rightarrow C>\dfrac{13}{18}\)