1*2022+2*2021+3*2020+.........................2022*1 tính
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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
\(B=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=-\dfrac{2}{3}\)
\(B=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=\dfrac{1}{3}-\dfrac{3}{3}=-\dfrac{2}{3}\)
Đặt
Đặt
Biểu thức có số số hạng là:
(số hạng)
Số nhóm được lập là:
(nhóm)
[ số hạng]
Vậy
\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)
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.
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