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

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

=>\(2a<\frac32\)

=>\(a<\frac34\)

b: \(M=1+3+3^2+\cdots+3^{99}+3^{100}\)

\(=\left(1+3\right)+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+\cdots+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+\cdots+3^{98}\left(1+3+3^2\right)\)

\(=4+13\left(3^2+3^5+\cdots+3^{98}\right)\)

=>M chia 13 dư 4

\(M=1+3+3^2+\cdots+3^{99}+3^{100}\)

\(=1+\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+\cdots+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=1+3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+\cdots+3^{97}\left(1+3+3^2+3^3\right)\)

\(=1+40\left(3+3^5+\cdots+3^{97}\right)\)

=>M chia 40 dư 1

a,       A = 1 + 3 + 3² +  3³ +....+3²⁰²²

     3A   =      3  + 3²  + 3³ +.....+3²⁰²² + 3²⁰²³

3A - A  =     3²⁰²³ - 1

      2A =     3²⁰²³ - 1

2A - 2²⁰²³ = 3²⁰²³ - 1 - 2²⁰²³ 

2A - 2²⁰²³ = -1 

b, x \(\in\) Z và x + 10 \(⋮\) x - 1 ( đk x# 1)

                      x + 10 \(⋮\) x - 1 

            \(\Leftrightarrow\) x - 1 + 11 \(⋮\) x - 1

                            11 \(⋮\) x - 1

                    x-1 \(\in\) { -11; -1; 1; 11}

                    x     \(\in\) { -10; 0; 2; 12}

Kết luận các số nguyên x thỏa mãn yêu cầu đề bài là :

                   x   \(\in\) { -10; 0; 2; 12}

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 + 3³ + 3⁴ + ... + 3²⁰²²

     3A = 3  + 3² + 3³ + ... + 3⁴ + ... + 3²⁰²² + 3²⁰²³

3A - A = (3 + 3² + 3³ + ... + 3⁴ + 3²⁰²² + 3²⁰²³) - (1 + 3+...+ 3²⁰²²)

2A     = 3 + 3² + 3³ + 3⁴ + ... + 3²⁰²² + 3²⁰²³ - 1 - 3 - ... - 3²⁰²²

2A =  (3 - 3) + (3² - 3²) + (3⁴ - 3⁴) + (3²⁰²² - 3²⁰²²) + (3²⁰²³ - 1)

2A = 3²⁰²³ - 1 

 A  = \(\dfrac{3^{2023}-1}{2}\)

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

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

B - A = \(\dfrac{3^{2023}}{2}\) - \(\dfrac{3^{2023}}{2}\) + \(\dfrac{1}{2}\)

B - A = \(\dfrac{1}{2}\)

Lồn bâm

Gâu gâu 

Bạn ko biết gõ số mũ à gõ thế này bố ai mà hiểu được

A=[1+3+3^2+3^3]+...+[3^2018+3^2019+3^2020+3^2021]

A=1 nhân[1+3+3^2+3^3]+...+3^2018 nhân [1+3+3^2+3^3]

A=[1+3+3^2+3^3] NHÂN[1+...+3^2018

A=40 nhân [1+...+3^2018]

=> A chia hết cho 40

Lời giải:
$A=1+(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2014}+3^{2015}+3^{2016})$

$=1+3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2014}(1+3+3^2)$
$=1+3.13+3^4.13+....+3^{2014}.13$

$=1+13(3+3^4+...+3^{2014})$ 

$\Rightarrow A-1\vdots 13(1)$

Mặt khác:
$A=1+(3+3^2+3^3+3^4)+....+(3^{2013}+3^{2014}+3^{2015}+3^{2016})$
$=1+3(1+3+3^2+3^3)+....+3^{2013}(1+3+3^2+3^3)$
$=1+(3+...+3^{2013})(1+3+3^2+3^3)$

$=1+40(3+....+3^{2013})$

$\Rightarrow A-1\vdots 5(2)$

Từ $(1); (2)$ mà $(5,13)=1$ nên $A-1\vdots (5.13)$ hay $A-1\vdots 65$

$\Rightarrow A$ chia $65$ dư $1$

em cảm ơn ạ