`# \text {DNamNgV}`

\(A=1+2+2^2+...+2^{2021}\text{ và }B=2^{2022}\)

Ta có:

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

Vì \(2^{2022}-1< 2^{2022}\)

\(\Rightarrow A< B.\)

A=B

a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.

A=1+3+3²+3³+.....+3²⁰²¹
-->3A=3(1+3+3²+3³+.....+3²⁰²¹)
-->3A=3+3²+3³+...+3²⁰²²
-->3A-A=(3+3²+3³+....3²⁰²²)-(1+3+3²+3³+.....+3²⁰²¹)
-->2A=3²⁰²²-1
-->A=(3²⁰²²-1):2
Vì (3²⁰²²-1):2>(3²⁰²²-1):2
-->A=B
 

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

\(2A=2^2+2^3+2^4+...+2^{2021}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2021}\right)-\left(2+2^2+2^3+...+2^{2021}\right)\)

\(A=2^2+2^3+2^4+...+2^{2021}-2-2^2-2^3-...-2^{2021}\)

\(A=2^{2021}-2\)

\(A=2+2^2+2^3+...+2^{2021}\\ \Leftrightarrow2A=2^2+2^3+2^4+...+2^{2022}\\ \Leftrightarrow2A-A=\left(2^2+2^3+2^4+...+2^{2022}\right)-\left(2+2^2+2^3+...+2^{2021}\right)\\ \Leftrightarrow A=2^{2022}-2\\ 2^{2022}-2< 2^{2022}\Rightarrow A< B\)

A = 2 + 2 2 + 2 3 + . . . + 2 2021 ⇔ 2 A = 2 2 + 2 3 + 2 4 + . . . + 2 2022 ⇔ 2 A − A = ( 2 2 + 2 3 + 2 4 + . . . + 2 2022 ) − ( 2 + 2 2 + 2 3 + . . . + 2 2021 ) ⇔ A = 2 2022 − 2 2 2022 − 2 < 2 2022 ⇒ A < B

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 

\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)

\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)

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

Ta có:

\(A=1+2+2^2+2^3+...+2^{2021}+2^{2022}\)

\(\Rightarrow2A=2\left(1+2+2^2+...+2^{2022}\right)\)

\(\Rightarrow2A=2+2^3+2^4+...+2^{2023}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2023}\right)-\left(1+2+2^2+...+2^{2022}\right)\)

\(\Rightarrow A=2^{2023}-1\)

Ta thấy: \(2^{2023}-1=2^{2023}-1\)

Vậy: \(A=B\)

\(A=2+2^2+...2^{2021}\)

\(\Rightarrow A+1=1+2+2^2+...2^{2021}\)

\(\Rightarrow A+1=\dfrac{2^{2021+1}-1}{2-1}\)

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

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

\(\Rightarrow A< B\)