Z=3¹+3²+3³+3⁴+...+3¹⁰⁰

3Z=3.(3¹+3²+3³+3⁴+...+3¹⁰⁰⁾

3Z=3.3¹+3.3²+3.3³+...+3.3¹⁰⁰

3Z=3²+3³+3⁴+...+3¹⁰¹

Lấy 3Z= 3²+3³+3⁴+...+3¹⁰¹     

 -

        Z=3¹+3²+3³+3⁴+...+3¹⁰⁰

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.

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

$\Rightarrow 3\vdots n+1$

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

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

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

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

Đặt A = 3¹ + 3² + 3³ + 3⁴ + ... + 3⁹⁹ + 3¹⁰⁰

= (3¹ + 3²) + (3³ + 3⁴) + ... + (3⁹⁹ + 3¹⁰⁰)

= 3.(1 + 3) + 3³.(1 + 3) + ... + 3⁹⁹.(1 + 3)

= 3.4 + 3³.4 + ... + 3⁹⁹.4

= 4.(3 + 3³ + ... + 3⁹⁹) ⋮ 4

Vậy A ⋮ 4

.

       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    = \(\dfrac{3^{101}+1}{4}\) 

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

       B  =  3¹ + 3² + 3³ +...+ 3¹⁰⁰

    3B   =         3² + 3³ + ...+ 3¹⁰⁰ + 3¹⁰¹

3B - B =      3¹⁰¹ - 3

2B     = 3¹⁰¹ - 3

2B + 3 = 3ⁿ

⇒ 3¹⁰¹   - 3 + 3= 3ⁿ

   3ⁿ = 3¹⁰¹

n = 101

Kết luận n = 101