1+2+3+...+2022=?
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\(2022A=2022+2022^2+2022^3+2022^4+...+2022^{2018}\)
\(2021A=2022A-A=2022^{2018}-1\Rightarrow A=\dfrac{2022^{2018}-1}{2021}\)
\(\Rightarrow A< B\)
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
-Ta có công thức với n∈N* thì:\(1+2+...+n=\dfrac{\left(\dfrac{n-1}{1}+1\right)\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)}{2}\)
\(B=1+\dfrac{1}{2}.\left(1+2\right)+\dfrac{1}{3}.\left(1+2+3\right)+...+\dfrac{1}{2022}.\left(1+2+3+...+2022\right)\)
\(=1+\dfrac{1}{2}.\dfrac{2.3}{2}+\dfrac{1}{3}.\dfrac{3.4}{2}+...+\dfrac{1}{2022}.\dfrac{2022.2023}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{2023}{2}\)
\(=\dfrac{2+3+4+...+2023}{2}=\dfrac{1+2+3+4+...+2022}{2}=\dfrac{\dfrac{2022.2023}{2}}{2}=10222626,5\)
\(1,\left(x+2022\right)\left(x-1\right)=x^2+2021x-2022\left(B\right)\\ 2,\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\left(A\right)\)
3a-b=1/2(a+b)
=>6a-2b=a+b
=>5a=3b
=>a/3=b/5=k
=>a=3k; b=5k
\(A=\dfrac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)
\(=\dfrac{3^{2022}\left(k^{2022}+1\right)}{5^{2022}\left(k^{2022}+1\right)}=\left(\dfrac{3}{5}\right)^{2022}\)
Ta có: \(\frac{2022}{2021^2+k}\le\frac{2022}{2021^2}\) (với \(k\)là số tự nhiên bất kì)
Ta có:
\(A=\frac{2022}{2021^2+1}+\frac{2022}{2021^2+2}+...+\frac{2022}{2021^2+2021}\)
\(\le\frac{2022}{2021^2}+\frac{2022}{2021^2}+...+\frac{2022}{2021^2}=\frac{2022}{2021^2}.2021=\frac{2022}{2021}\)
Ta có: \(\frac{2022}{2021^2+k}>\frac{2022}{2021^2+2021}=\frac{2022}{2021.2022}=\frac{1}{2021}\)với \(k\)tự nhiên, \(k< 2021\))
Suy ra \(A=\frac{2022}{2021^2+1}+\frac{2022}{2021^2+2}+...+\frac{2022}{2021^2+2021}\)
\(>\frac{1}{2021}+\frac{1}{2021}+...+\frac{1}{2021}=\frac{2021}{2021}=1\)
Suy ra \(1< A\le\frac{2022}{2021}\)do đó \(A\)không phải là số tự nhiên.
Số số hạng là : ( 2022 - 1 ) : 1 + 1 = 2022 ( số )
Tổng là : ( 2022 + 1 ) x 2022 : 2 = 2045253
Đ/S:...
1+2+3+....+2022=2045253