(1 - 1/ 1+2).(1 - 1/ 1+2+3).....(1 - 1/ 1+2+3+...+2022), so sánh với 1/2
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MM
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 5 2023
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
17 tháng 5 2022
\(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2022^2}\)
\(C=\dfrac{1}{\left(2+3+4+...+2022\right)^2}\)
\(C=\dfrac{1}{\left(2022-2+1\right)^2}\)
\(C=\dfrac{1}{2021^2}\)
\(C=\dfrac{1}{2021\cdot2021}\)
\(C=\dfrac{1\div2021}{2021}\)
\(C=1\)
Vì \(1>\dfrac{13}{18}\)
\(\Rightarrow C>\dfrac{13}{18}\)
23 tháng 3 2022
\(1+2+...+n=\dfrac{\left(\dfrac{n-1}{1}+1\right).\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)}{2}\)
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=3\left(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+2022}\right)\)
\(=3\left(\dfrac{1}{\dfrac{2.\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3.\left(3+1\right)}{2}}+...+\dfrac{1}{\dfrac{2022.\left(2022+1\right)}{2}}\right)\)
\(=3\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2022.2023}\right)\)
\(=3.2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)\)
\(=6.\dfrac{2021}{4046}=3.\dfrac{2021}{2023}=\dfrac{6063}{2023}=\dfrac{18189}{6069}\)
\(\dfrac{10}{3}=\dfrac{20230}{6069}>\dfrac{18189}{6069}=M\)
9 tháng 1
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)
\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)
\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)
\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)
mà \(3< \dfrac{10}{3}\)
nên \(M< \dfrac{10}{3}\)
BD
16 tháng 7 2023
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
GB
1
GB
0
NT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
9 tháng 5 2022
\(2022A=2022+2022^2+2022^3+2022^4+...+2022^{2018}\)
\(2021A=2022A-A=2022^{2018}-1\Rightarrow A=\dfrac{2022^{2018}-1}{2021}\)
\(\Rightarrow A< B\)
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