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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
 

AH
Akai Haruma
Giáo viên
22 tháng 5 2023

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

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

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

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

$\Rightarrow A< B$

22 tháng 5 2023

`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`

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 

 

20 tháng 10 2023

A=2+22+23+...+22021

2A=22+23+24+...+22022

2A-A=(22+23+24+...+22022)-(2+22+23+...+22021)

A=22022-2 mà B= 22022 nên A<B.

6 tháng 5 2022

a) \(A=2A-A\)

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

\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

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

b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)

\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)

 

6 tháng 5 2022

a) A = 2 A − A = 2 ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 + 1 2 + . . . + 1 2 2021 − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 − 1 2 2022 b) B = 20 + 15 + 12 + 17 60 = 4 5 = 1 − 1 5 A > B ( V ì ( 1 2 2022 < 1 5 ) )

28 tháng 12 2021

vuigiúp mk vs

28 tháng 12 2021

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

\(\Rightarrow2a=2+2^2+2^3+...+2^{2022}\)

\(\Rightarrow2a-a=2+2^2+2^3+...+2^{2022}-1-2-2^2-...-2^{2021}\)

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

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

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

3 tháng 12 2021

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

17 tháng 10 2021

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

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

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

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

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