S = (1 - 3 + 3² - 3³) + 3⁴ . (1 - 3 + 3² - 3³) + .... + 3⁹⁶ . (1 - 3 + 3² - 3³)

S = (-20) + 3⁴ . (-20) +.... + 3⁹⁶ . (-20)

S = (-20) . (1 + 3⁴ +...+ 3⁹⁶) 

\(\Rightarrow\)S \(⋮\) 20 

(Ko bt có đúng ko)

*KO CHÉP MẠNG*

\(S=1-3+3^2-3^3+...+3^{98}-3^{99}=\left(1-3+3^2-3^3\right)+3^4\left(1-3+3^3-3^3\right)+...+3^{96}\left(1-3+3^2-3^3\right)=\left(-20\right)+3^4.\left(-20\right)+...+3^{96}.\left(-20\right)=\left(-20\right)\left(1+3^4+...+3^{96}\right)⋮20\)

Ta có: \(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)

\(=\left(1-3+3^2-3^3\right)+...+3^{96}\left(1-3+3^2-3^3\right)\)

\(=-20\cdot\left(1+...+3^{96}\right)⋮20\)

\(S=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)

\(=13+3^3.13+...+3^{96}.13=13\left(1+3^3+...+3^{96}\right)⋮13\)

  a,

S  =     1 -  3 + 3² - 3³+...+3⁹⁸ - 3⁹⁹

S  =   3⁰ - 3¹ + 3² - 3³+...+ 3⁹⁸ - 3⁹⁹

xét dãy số: 0; 1; 2; 3;...;99 

Dãy số trên là dãy số cách đều với khoảng cách là: 1 - 0 = 1

Dãy số trên có số số hạng là: (99 - 0): 1 + 1 = 100 (số)

100 : 4 = 25

Vậy ta nhóm 4 số hạng liên tiếp của tổng S thành 1 nhóm thì: 

S = ( 1 - 3 + 3² - 3³) +....+( 3⁹⁶ - 3⁹⁷ + 3⁹⁸ - 3⁹⁹)

S = - 20+...+ 3⁹⁶.(1 - 3 + 3² - 3³)

S = - 20 +...+ 3⁹⁶.(-20)

S = -20.( 3⁰ + ...+ 3⁹⁶) (đpcm)

b,

  S           = 1 - 3 + 3² - 3³+...+ 3⁹⁸ - 3⁹⁹

3S          =      3  - 3² + 3³-...-3⁹⁸  + 3⁹⁹ - 3¹⁰⁰

3S + S   =    - 3¹⁰⁰ + 1

4S        = - 3¹⁰⁰ + 1 

 S        = ( -3¹⁰⁰ + 1): 4

S        = - ( 3¹⁰⁰ - 1) : 4

Vì S là số nguyên nên 3¹⁰⁰ - 1 ⋮ 4 ⇒ 3¹⁰⁰ : 4 dư 1 (đpcm)

`#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

\(A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)

\(=13+3^3.13+...+3^{96}.13\)

\(=13\left(1+3^3+...+3^{96}\right)⋮13\)

\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ A=13\left(1+3^3+...+3^{96}\right)⋮13\)

mn ơi mong mọi người trả lời câu hỏi giúp mik với ạ 

Bài 1:

a. $2^{29}< 5^{29}< 5^{39}$

$\Rightarrow A< B$

b.

$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$

$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$

$=(1+3)(3+3^3+3^5+...+3^{2009})$

$=4(3+3^3+3^5+...+3^{2009})\vdots 4$

Mặt khác:

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$

$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$

Bài 1:
c.

$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$

$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$

$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$

$\Rightarrow A=\frac{3^{101}+1}{4}$

Bài 2:

a. $7\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 7; -7\right\}$

$\Rightarrow n\in \left\{0; -2; 6; -8\right\}$

b.

$2n+5\vdots n+1$
$\Rightarrow 2(n+1)+3\vdots n+1$

$\Rightarrow 3\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow n\in \left\{0; -2; 2; -4\right\}$

a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)

\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)

\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)

\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)

\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)

\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)

\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)

\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)