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Giả sử tất cả các số đã cho đều lẻ
=>Quy đồng, ta được:
\(A=\dfrac{\left(a_2\cdot a_3\cdot...\cdot a_{2022}\right)+\left(a_1\cdot a_3\cdot...\cdot a_{2021}\cdot a_{2022}\right)+...+\left(a_1\cdot a_2\cdot...\cdot a_{2021}\right)}{a_1\cdot a_2\cdot...\cdot a_{2022}}=1\)
Tử có 2022 số hạng, mẫu là số lẻ
=>A là số chẵn khác 1
=>Trái GT
=>Phải có ít nhất 1 số là số chẵn
C = 1 - 2 + 22 - 23 + 24 -...+ 22022
2C = 2 - 22 + 23 - 24 +...- 22022 + 22023
2C + C = 22023 + 1
3C = 22023 + 1
C = \(\dfrac{2^{2023}+1}{3}\)
\(S=\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2022}}\)
=>\(25\cdot S=1+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2020}}\)
=>\(25S-S=1+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2020}}-\dfrac{1}{5^2}-\dfrac{1}{5^4}-...-\dfrac{1}{5^{2022}}\)
=>\(24S=1-\dfrac{1}{5^{2022}}\)
=>\(S=\dfrac{1}{24}-\dfrac{1}{24\cdot5^{2022}}< \dfrac{1}{24}\)
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=1/2^2+1/3^2+...+1/2022^2<1-1/2+1/2-1/3+...+1/2021-1/2022<1
mà A>0
nên 0<A<1
=>A ko là số tự nhiên
\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2022}{4^{2022}}\)
\(\Rightarrow S< \dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{2}{4^3}+...+\dfrac{1}{4^{2021}}\)
\(\Rightarrow S< \dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{2}{4^3}+...+\dfrac{1}{4^{2021}}\left(1\right)\)
Đặt \(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{2}{4^3}+...+\dfrac{1}{4^{2021}}\)
\(\Rightarrow\dfrac{1}{4}A=\left(\dfrac{1}{4^2}+\dfrac{1}{4^3}+\dfrac{2}{4^4}+...+\dfrac{1}{4^{2022}}\right)\)
\(\Rightarrow A-\dfrac{1}{4}A=\dfrac{1}{4}-\dfrac{1}{4^{2022}}=\dfrac{1}{4}\left(1-\dfrac{1}{4^{2021}}\right)\)
\(\left(1\right)\Rightarrow S< \dfrac{1}{4}+\dfrac{1}{4}\left(1-\dfrac{1}{4^{2021}}\right)< \dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
Vậy \(S< \dfrac{1}{2}\left(đpcm\right)\)