a: \(A=1+3+3^2+\cdots+3^{2022}\)

=>\(3A=3+3^2+3^3+\cdots+3^{2023}\)

=>\(3A-A=3+3^2+\cdots+3^{2023}-1-3^{}-\cdots-3^{2022}\)

=>\(2A=3^{2023}-1\)

=>\(2A-3^{2023}=-1\)

b: x+10⋮x-1

=>x-1+11⋮x-1

=>11⋮x-1

=>x-1∈{1;-1;11;-11}

=>x∈{2;0;12;-10}

A =1+3+3²+.....+3²⁰²²+3²⁰²³

3.A =3+3²+3³+.....+3²⁰²³+3²⁰²⁴

3.A -A=(3+3²+3³+.....+3²⁰²³+3²⁰²⁴ ) - (1+3+3²+.....+3²⁰²²+3²⁰²³)

2A =3²⁰²⁴-1

A =\(\dfrac{3^{2024}-1}{2}\)

A = 1 + 3 + 3² + ... + 3²⁰²³

⇒ 3A = 3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴

⇒ 2A = 3A - A

= (3 + 3² + 3³ + ... + 3²⁰²³ + 3²⁰²⁴) - (1 + 3 + 3² + ... + 3²⁰²³)

= 3²⁰²⁴ - 1

⇒ A = (3²⁰²⁴ - 1) : 2

⇒ A < B

A=1+3+3²+3³+3⁴+........+3²⁰²²+3²⁰²³

3A=3+3²+3³+............+3²⁰²³+3²⁰²⁴

3A-A=(3+3²+3³+..........+3²⁰²³+3²⁰²⁴

\(S=3^{2024}-3^{2023}+3^{2022}-3^{2021}+...+3^2-3\)

\(3S=3^{2025}-3^{2024}+3^{2023}-3^{2022}+...+3^3-3^2\)

\(3S+S=3^{2025}-3^{2024}+3^{2023}-3^{2022}+...+3^3-3^2+3^{2024}-3^{2023}+3^{2022}-3^{2021}+...+3^2-3\)\(4S=3^{2025}-3\)

\(S=\dfrac{3^{2025}-3}{4}\)

         S = 3²⁰²⁴ - 3²⁰²³ + 3²⁰²² - 3²⁰²¹ +... + 3² - 3

      3.S = 3²⁰²⁵ - 3²⁰²⁴ + 3²⁰²² -3²⁰²¹ + ....+ 3³ - 3²

3S + S = 3²⁰²⁵ - 3²⁰²⁴ + 3²⁰²² - 3²⁰²¹ +...+3³ - 3²+(3²⁰²⁴-3²⁰²³+...-3)

   4S    = 3²⁰²⁵ - 3²⁰²⁴ + 3²⁰²² - 3²⁰²¹+...+3³-3² + 3²⁰²⁴-3²⁰²³+...-3

    4S = 3²⁰²⁵ - (3²⁰²⁴ - 3²⁰²⁴) -...-(3² - 3²) - 3

    4S = 3²⁰²⁵ - 3

      S = \(\dfrac{3^{2025}-3}{4}\)

Lời giải:

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

$3A=3+3^2+3^3+...+3^{2022}$

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

$\Rightarrow 2A=3^{2022}-1$

$\Rightarrow A=\frac{3^{2022}-1}{2}$

$B-A=\frac{3^{2022}}{2}-\frac{3^{2022}-1}{2}=\frac{1}{2}$

Ta có: \(a=\frac13+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\cdots+\frac{2022}{3^{2022}}\)

=>\(3a=1+\frac23+\frac{3}{3^2}+\frac{4}{3^3}+\cdots+\frac{2022}{3^{2021}}\)

=>\(3a-a=1+\frac23+\frac{3}{3^2}+\cdots+\frac{2022}{3^{2021}}-\frac13-\frac{2}{3^2}-\frac{3}{3^3}-\cdots-\frac{2022}{3^{2022}}\)

=>\(2a=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{2021}}-\frac{2022}{3^{2022}}\)

Đặt \(b=\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{2021}}\)

=>\(3b=1+\frac13+\cdots+\frac{1}{3^{2020}}\)

=>\(3b-b=1+\frac13+\ldots+\frac{1}{3^{2020}}-\frac13-\frac{1}{3^2}-\cdots-\frac{1}{3^{2021}}\)

=>\(2b=1-\frac{1}{3^{2021}}=\frac{3^{2021}-1}{3^{2021}}\)

=>\(b=\frac{3^{2021}-1}{2\cdot3^{2021}}\)

Ta có: \(2a=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{2021}}-\frac{2022}{3^{2022}}\)

=>\(2a=1+\frac{3^{2021}-1}{2\cdot3^{2021}}-\frac{2022}{3^{2022}}=1+\frac{3^{2022}-3-4044}{2\cdot3^{2022}}=1+\frac12-\frac{4047}{2\cdot3^{2022}}\)

=>\(2a<\frac32\)

=>\(a<\frac34\)

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

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

\(=1+3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2020}\left(1+3+3^2\right)\)

\(=1+13\left(3+3^4+...+3^{2020}\right)\)

=>A chia 13 dư 1

Bạn ơi, bạn cũng xem lại giúp mình luôn nha

2020 đâu có chia hết cho 3

Với lại dãy này có 2023 số đó bạn, 2023 cũng đâu chia hết cho 3 đâu

Trường nào đó?

Bài 1

a) S = 1 + 2 + 2² + 2³ + ... + 2²⁰²³

2S = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²⁴

S = 2S - S = (2 + 2² + 2³ + ... + 2²⁰²⁴) - (1 + 2 + 2² + 2³)

= 2²⁰²⁴ - 1

b) B = 2²⁰²⁴

B - 1 = 2²⁰²⁴ - 1 = S

B = S + 1

Vậy B > S

a,

\(S=1+2+2^2+...+2^{2023}\)

\(2S=2+2^2+2^3+...+2^{2024}\)

\(\Rightarrow S=2^{2024}-1\)

b.

Do \(2^{2024}-1< 2^{2024}\)

\(\Rightarrow S< B\)

2.

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

\(\Rightarrow3H=3^2+3^3+...+3^{2023}\)

\(\Rightarrow3H-H=3^{2023}-3\)

\(\Rightarrow2H=3^{2023}-3\)

\(\Rightarrow H=\dfrac{3^{2023}-3}{2}\)