Question

cho A=1+3+3²+...+3²⁰²³ va B=3²⁰²⁴:2 tinh (2B-2A)²⁰²³+2022

Answer 1

Lời giải:

$A=1+3+3^2+...+3^{2023}$
$3A=3+3^2+3^3+....+3^{2024}$

$\Rightarrow 3A-A=3^{2024}-1$

$\Rightarrow 2A=3^{2024}-1$
$B=3^{2024}:2$
$\Rightarrow 2B=3^{2024}$

Suy ra:

$(2B-2A)^{2023}+2022=[3^{2024}-(3^{2024}-1)]^{2023}+2022$
$=1^{2023}+2022=2023$

Answer 2

`A = 1 +3 + ... + 3^2023`

`=> 3A = 3 + 3^2 + ... +3^2024`

`=> 3A - A = 3^2024 - 1`

`=> 2A = 3^2024 - 1`

Có : `B = 3^2024: 2`

`=> 2B = 3^2024`

`=> (2B - 2A)^2023 + 2022`

`= (3^2024 - 3^2024 + 1)^2023 + 2022`

`= 1^2023 + 2022`

`= 1 + 2022`

`=2023`