B=1*2*3*...*2021-1*2*3*...*2020-1*2*3*...*2019*2020^2 Giá trị của B là? giúp mình nốt câu này với ạ
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Những câu hỏi liên quan
15 tháng 6 2023
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 12 2023
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
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.
NP
4
10 tháng 9 2020
số lượng số hạng của dãy số là
( 2021 - 2 ) : 1 + 1 = 2020
tổng của dãy số là
( 2021 + 2) x 2020 : 2 = 2043230
vậy A = \(\frac{1}{2043230}\)
SI
29 tháng 6 2021
Ta có :
B = \(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
B = \(\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)
B = \(\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+1\)
B = \(2021\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+...+\dfrac{1}{2}\right)\) (1)
Mà A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{A}{B}=\dfrac{1}{2021}\)
29 tháng 6 2021
Ta có: \(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
\(=\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)
\(=\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+\dfrac{2021}{2021}\)
Suy ra: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)