Question

ChoA=1+2^1+2^2+2^3+...+2^2020+2^2021

a Tính 2.A

b chứng minhA=2^2022-1

Answer 1

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

Answer 2

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

Answer 3

a/

\(2A=2+2^2+2^3+...+2^{2022}\)

b/

\(A=2A-A=2^{2022}-1\)

Answer 4

a) \(A=1+2^1+2^2+2^3+...+2^{2020}+2^{2021}\)

\(\Rightarrow2A=2\left(1+2^1+2^2+2^3+...+2^{2020}+2^{2021}\right)\)

\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2020}+2^{2021}+2^{2022}\)

b) Ta có: \(2A=2+2^2+2^3+...+2^{2021}+2^{2022}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2021}+2^{2022}\right)-\left(1+2+2^2+...+2^{2020}+2^{2021}\right)\)

\(\Rightarrow A=\left(2-2\right)+\left(2^2-2^2\right)+\left(2^3-2^3\right)+...+\left(2^{2022}-1\right)\)

\(\Rightarrow A=2^{2022}-1\)

Answer 5

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