Nếu đúng là zậy thì mk biết làm.

A = 3 + 3² + 3³ + ...  + 3²⁰⁰⁴

A = (  3 + 3² + 3³ + 3⁴ ) + ... + ( 3²⁰⁰¹ + 3²⁰⁰² + 3²⁰⁰³ + 3²⁰⁰⁴ )

A = 3( 1 + 3 + 3² + 3³ ) + ... + 3²⁰⁰¹( 1 + 3 + 3² + 3⁹ )

A = 3.40 + ... + 3²⁰⁰¹.40

A = ( 3 + 3⁵ + ...  3²⁰⁰¹) . 40

=> A chia hết cho 40

A = 3 + 3² + 3³ +3⁴ + ... + 3²⁰⁰⁴ phải ko? 

sao lai la 3^9 vay bn

Tham khảo

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)

2A = 3101−13101−1

⇒⇒ A = 3101−123101−12

Vậy A = 3101−12

A=3 mũ 101-1 phân số2

\(A=1-3+3^2-3^3+3^4-...-3^{98}-3^{99}+3^{100}\\ 3A=3-3^2+3^3-3^4-...-3^{98}+3^{99}-3^{100}+3^{101}\\ 3A-A=3^{101}-1\\ \Rightarrow A=\dfrac{3^{101}-1}{2}\)

       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 = \(\dfrac{1}{2}\)

       A =  1 - 3 +  3² -   3³ + 3⁴ - ... + 3⁹⁸ - 3⁹⁹ + 3¹⁰⁰

      3A =  3 - 3² + 3³ - 3⁴+ 3⁵  - ... + 3⁹⁹ - 3¹⁰⁰ + 3¹⁰¹

3A + A = 3 - 3²+ 3³-3⁴+3⁵ -...+3⁹⁹ - 3¹⁰⁰ + 3¹⁰¹ + 1 - 3 +...-3⁹⁹+3¹⁰⁰

4A   =    3¹⁰¹ + 1

  A    = \(\dfrac{3^{101}+1}{4}\) 

Lời giải:
$T=3-3^2+3^3-3^4+....-3^{2000}$

$3T=3^2-3^3+3^4-3^5+...-3^{2001}$

$\Rightarrow T+3T=3-3^{2001}$

$\Rightarrow 4T=3-3^{2001}$

$\Rightarrow T=\frac{3-3^{2001}}{4}$

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.

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

$\Rightarrow 3\vdots n+1$

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

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