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A = \(\dfrac{\dfrac{2022}{1}+\dfrac{2021}{2}+\dfrac{2020}{3}+...+\dfrac{1}{2022}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}}\)
Xét TS = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) \(\dfrac{2020}{3}\) +... + \(\dfrac{1}{2022}\)
TS = (1 + \(\dfrac{2021}{2}\)) + (1 + \(\dfrac{2020}{3}\)) + ... + ( 1 + \(\dfrac{1}{2022}\)) + 1
TS = \(\dfrac{2023}{2}\) + \(\dfrac{2023}{3}\) +...+ \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2023}\)
TS = 2023.(\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) +...+ \(\dfrac{1}{2023}\))
A = \(\dfrac{2023.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}\)
A = 2023
a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)
b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)
c) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
Lời giải:
$A=1-3+3^2-3^3+...+3^{2022}-\frac{3^{2023}}{4}$
$3A=3-3^2+3^3-3^4+...+3^{2023}-\frac{3^{2024}}{4}$
$\Rightarrow A+3A=1+3^{2023}-\frac{3^{2023}}{4}-\frac{3^{2024}}{4}$
$\Rightarrow 4A=1$
$\Rightarrow A=\frac{1}{4}$
Ta có \(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\)
\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\)\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\)
Đặt B = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)
2B = \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)
2B - B = \(\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)B = 2 - \(\dfrac{1}{2^{2022}}\)
Suy ra A = 2 - \(\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\) < 2
Vậy A < 2
\(A=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{3}{2^{3}}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2\)
Có:
\(1+2+2^2+2^3+...+2^{2022}\)
\(=\left(2+2^3+2^4+...+2^{2023}\right)-\left(1+2+2^2+2^3+...+2^{2022}\right)\)
\(=2^{2023}-1\)
\(\Rightarrow A=\dfrac{2^{2023}-1}{1-2^{2023}}=-1\)