Question

[boi đz]

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2022}}\) và \(B=1-\dfrac{1}{3^{2021}}\)

So sánh A và B

Answer 1

[Akai Haruma]

Lời giải:
$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2022}}$

$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2021}}$

$\Rightarrow 3A-A=1-\frac{1}{3^{2022}}$

$\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{2022}}$

Xét hiệu:
$A-B=\frac{1}{2}-\frac{1}{2.3^{2022}}-(1-\frac{1}{3^{2021}})$

$=\frac{1}{3^{2021}}-\frac{1}{2.3^{2022}}-\frac{1}{2}$

$=\frac{5}{2.3^{2022}}-\frac{1}{2}$

$< \frac{1}{2}-\frac{1}{2}=0$

$\Rightarrow A< B$

Answer 2

[TV Cuber]

`A = 1/3 +1/3^2 +1/3^3 +...+1/3^2022`

`<=> 3A = 1 +1/3 +1/3^2 +...+ 1/3^2021`

`=>2A =3A-A =1+1/3 +1/3^2 +..+ 1/3^2021 - 1/3-1/3^2-1/3^3..-1/3^2022`

`2A = 1-1/3^2022`

`=> A = (1-1/3^2022) :2`

Ta thấy `1- 1/3^2022 < 1-1/3^2021`

`=> (1 -1/3^2022):2<1-1/3^2021`

Hay `A<B`