ChoA=1+2^1+2^2+2^3+...+2^2020+2^2021
a Tính 2.A
b chứng minhA=2^2022-1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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.
15 tháng 6 2023
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
LP
2
24 tháng 2 2022
\(2A=\dfrac{2^{2021}-1-1}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)
\(2B=\dfrac{2^{2022}-1-1}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)
mà \(2^{2021}-1< 2^{2022}-1\)
nên A<B
27 tháng 2 2022
A=22020−122021−1A=22020-122021-1
⇒2A=2.(22020−1)22021−1⇒2A=2.(22020-1)22021-1
⇒2A=22021−222021−1⇒2A=22021-222021-1
⇒2A=22021−1−122021−1⇒2A=22021-1-122021-1
⇒2A=1−122021−1⇒2A=1-122021-1
B=22021−122022−1B=22021-122022-1
⇒2B=2.(22021−1)22022−1⇒2B=2.(22021-1)22022-1
13 tháng 7 2023
\(B=\left(\dfrac{2020}{2}+1\right)+\left(\dfrac{2019}{3}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1\)
\(=\dfrac{2022}{2}+\dfrac{2022}{3}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}\)
=2022(1/2+1/3+...+1/2021+1/2022)
=>B/A=2022
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
25 tháng 2 2022
\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)
\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)
dó \(\frac{1}{2^{2022}-1}< \frac{1}{2^{2021}-1}\Rightarrow1-\frac{1}{2^{2022}-1}>1-\frac{1}{2^{2021}-1}\Rightarrow A< B\)
HT
5N
3
8 tháng 10 2023
Đặt B=2023−2022+2021−2020+...+3−2+1�=2023-2022+2021-2020+...+3-2+1
B=(2023−2022)+(2021−2020)+...+(3−2)+1�=(2023-2022)+(2021-2020)+...+(3-2)+1
Đặt A=(2023−2022)+(2021−2020)+...+(3−2)�=(2023-2022)+(2021-2020)+...+(3-2)
Biểu thức A� có số số hạng là:
(2023−2):1+1=2022(2023-2):1+1=2022 (số hạng)
Số nhóm được lập là:
2022:2=10112022:2=1011 (nhóm)
A=1+1+...+1�=1+1+...+1 [10111011 số hạng]
A=1×1011=1011�=1×1011=1011
⇒B=1011+1=1012⇒�=1011+1=1012
Vậy B=1012
a) Ta có A = 1 + 21 + 22 + ... + 22021
2A = 21 + 22 + 23 + ... + 22022
Vậy 2A = 21 + 22 + 23 + ... + 22022
b) 2A - A = ( 21 + 22 + 23 + ... + 22022 ) - ( 1 + 21 + 22 + ... + 22021 )
A = 22022 - 1
Vậy A = 22022 - 1
a)
\(A=1+2^1+2^2+2^3+...+2^{2020}+2^{2021}\)
\(2A=2^1+2^2+2^3+2^4+...+2^{2021}+2^{2022}\)
b)
\(2A=2^1+2^2+2^3+...+2^{2022}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2022}\right)-\left(1+2^1+2^2+....+2^{2021}\right)\)
\(A=2^{2022}-1\)
=> đpcm