Dịch ra là: Ta có: 3A = 3. (1 + 3 + 32 + 33 + ... + 399 + 3100) (1 + 3 + 32 + 33 + ... + 399 + 3100) 3A = 3 + 32 + 33 + ... + 3100 + 31013 + 32 + 33 + ... + 3100 + 3101 Suy ra: 3A - A = (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) ⇒⇒ A = 3101−123101−12 Vậy A = 3101−12

Mà đoạn 2A sai nhé bạn, sửa lại:

2A = 3101−13101−1 2A=-10001

A=-10001/2

A=-5000,5

Vậy A=-5000,5

xin lỗi bài trên của mình làm sai

Ta có: 3A = 3.(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰⁾ 

3A = 3+3²+3³+...+3¹⁰⁰+3¹⁰¹

Suy ra: 3A – A = (3+3²+3³+...+3¹⁰⁰+3¹⁰¹)−(1+3+3²+3³+...+3⁹⁹+3¹⁰⁰)

2A = 3¹⁰¹−1

⇒ A = 3¹⁰¹−1

             2               

Vậy A = 3¹⁰¹−1

                 2           

em den lam

\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)

A=3²⁰¹⁹+1+3+3²+3³+...+3²⁰¹⁸

⇒A=1+3+3²+...+3²⁰¹⁸+3²⁰¹⁹ 

⇒3A=3×(1+3+3^2+3^3+....+3^2019)

3A=3+3^2+3^3+....+3^2020

3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)

2A= 3^2020-1

⇒ A =( 3^2020-1):2

A=3²⁰¹⁹+1+3+3²+3³+...+3²⁰¹⁸

⇒A=1+3+3²+...+3²⁰¹⁸+3²⁰¹⁹ 

⇒3A=3×(1+3+3^2+3^3+....+3^2019)

⇒3A=3+3^2+3^3+....+3^2020

⇒3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)

⇒2A= 3^2020-1

⇒ A =( 3^2020-1):2

Số các số hạng là: 101 – 0 + 1 = 102 số.
Ta nhận thấy:
1 + 3 + 32 = 1 + 3 + 9 = 13;
33 + 34 + 35 = 33(1 + 3 + 32) = 33.13;
…
Mà 102 có tổng các chữ số là 1 + 0 + 2 = 3 chia hết cho 3 nên 102 chia hết cho 3, nghĩa là:
A = (1 + 3 + 32) + (33 + 34 + 35) + … + (399 + 3100 + 3101)
= (1 + 3 + 32) + 33(1 + 3 + 32) + … + 399(1 + 3 + 32)
= 13 + 33.13 + … + 399.13
= 13.(1 + 33 + … + 399) chia hết cho 13.
Vậy A chia hết cho 13.

  A = 1 +  3  + 3² + 3³ + ... + 3¹⁰⁰

3A = 3 + 3² + 3³ +3⁴+ .... + 3¹⁰¹

3A - A = (3 + 3² + 3⁴ + ... + 3¹⁰¹) - (1 + 3 + 3² + 3³ + ... + 3¹⁰⁰)

2A     = 3 + 3² + 3⁴ + ... + 3¹⁰¹ - 1 - 3 - 3² - 3³ - ... - 3¹⁰⁰

2A = (3 - 3) + (3² - 3²) + ... + (3¹⁰⁰ - 3¹⁰⁰) + (3¹⁰¹ - 1)

2A = 3¹⁰¹ - 1

A = \(\dfrac{3^{101}-1}{2}\)

\(A=1+3+3^2+3^3+...+3^{2018}+3^{2019}\)

\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{2018}\left(1+3\right)\)

\(=\left(1+3\right)\left(1+3^2+...+3^{2018}\right)\)

\(=4\left(1+3^2+...+3^{2018}\right)\) ⋮4

⇒A⋮4

Lời giải:

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

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

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

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

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

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

`#3107.101107`

\(A = 1 + 3 + 3^2 + 3^3 + ... + 3^{98} + 3^{99}\)

\(A = (1 + 3) + (3^2 + 3^3) + ... + (3^{98} + 3^{99})\)

\(A = (1 + 3) + 3^2(1 + 3) + ... + 3^{98}(1 + 3)\)

\(A = (1 + 3)(1 + 3^2 + ... + 3^{98})\)

\(A = 4(1 + 3^2 + ... + 3^{98})\)

Vì \(4(1 + 3^2 + ... + 3^{98}) \) \(\vdots\) \(4\)

`\Rightarrow A \vdots 4`

Vậy, `A \vdots 4` (đpcm).

A = 1 + 3 + 3² + 3³ + ... + 3⁹⁸ + 3⁹⁹

A = (1 + 3) + (3² + 3³) + ... + (3⁹⁸ + 3⁹⁹)

A = 1. (1 + 3) + 3². (1 + 3) + ... + 3⁹⁸. (1 + 3)

A = 1.4 + 3².4 + ... + 3⁹⁸.4

A = 4. (1 + 3² + ... + 3⁹⁸)

⇒ A ⋮ 4

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