\(A=1+3+3^2+...+3^{2016}+3^{2017}\)

\(3A=3+3^2+3^3+...+3^{2017}+3^{2018}\)    

\(3A-A=3^{2018}-1\)

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

Vậy n = 2018

3A=3+3^2+3^3+...+3^2018

-A=1+3+3^2+...+3^2017

2A=3^2018-1

khi đó ta có 2A+1=3^2018-1+1=3^2018=3^n

=>n=2018

cảm ơn các bạn

nhanh tích cho nhee

tui làm b nha do a không biết làm

A=5+3²+3³+...+3²⁰¹⁸

3A=15+3³+3⁴+...+3²⁰¹⁹

3A-A=(15+3³+3⁴+...+3²⁰¹⁹)-(5+3²+3³+...+3²⁰¹⁸)

2A=3²⁰¹⁹+15-(5+3²)

2A=3²⁰¹⁹+15-14

2A=3²⁰¹⁹+1

2A-1=3²⁰¹⁹+1-1

2A-1=3²⁰¹⁹

vậy n = 2019

cmon nhaa, mỗi câu b thoi cx đc :3

OLM duyệt nhanh lên nhé!

ta có A=1+3+3²+3³+......+3⁹⁹+3¹⁰⁰

=>3A= 3+3²+3³+3⁴+......+3¹⁰⁰+3¹⁰¹

- A=1+3+3²+3³+.......+3⁹⁹+3¹⁰⁰

=> 2A=3¹⁰¹-1 mà 2A+1=3ⁿ =>3¹⁰¹-1+1

                                           => 3¹⁰¹-3ⁿ

                                           => n= 101

k cho mik nha!

n=101 k để mik nên điểm nha^_^

mn mn ơiii

helllppppppppp

\(2,\\ 3^{n-3}+2^{n-3}+3^{n+1}+2^{n+2}\\ =3^{n-3}\left(1+3^4\right)+2^{n-3}\left(1+2^5\right)\\ =3^{n-3}\cdot82+2^{n-3}\cdot33\)

Vì \(3^{n-3}\cdot82⋮2;⋮3\) nên \(3^{n-3}\cdot82⋮6\)

\(2^{n-3}\cdot33⋮2;⋮3\) nên \(2^{n-3}\cdot33⋮6\)

Do đó tổng trên chia hết cho 6 với mọi \(n\in N\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{11}\\ \Rightarrow3A-A=\left(3+3^2+...+3^{11}\right)-\left(1+3+...+3^{10}\right)\\ \Rightarrow2A=3^{11}-1\\ \Rightarrow2A+1=3^{11}=3^n\\ \Rightarrow n=11\)

\(A=3+3^2+...+3^{2008}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2009}\)

\(\Rightarrow3A-A=3^{2009}-3\)

\(\Rightarrow2A+3=3^{2009}\)

Vậy n = 2009

\(A=3+3^2+3^3+...+3^{2008}\)

\(\Leftrightarrow3A=3^2+3^3+...+3^{2009}\)

\(\Leftrightarrow3A-A=3^{2009}-3\Leftrightarrow2A+3=3^{2009}\)

Vậy n=2009

a, 10 ⋮ 3a+1 => 3a+1 ∈ Ư(10) => 3a+1 ∈ {1;2;5;10} => a ∈ { 0 ; 1 3 ; 4 3 ; 3 }. Vì a ∈ N, a ∈ {0;3}

b, a+6 ⋮ a+1 => a+1+5 ⋮ a+1 => 5 ⋮ a+1 => a+1 ∈ Ư(5) =>  a+1 ∈ {1;5} => a ∈ {0;4}

c, 3a+7 ⋮ 2a+3 => 2.(3a+7) - 3(2a+3) ⋮ 2a+3 => 5 ⋮ 2a+3 => 2a+3 ∈ Ư(5)

=> 2a+3 ∈ {1;5} => a = 1

d, 6a+11 ⋮ 2a+3 => 3.(2a+3)+2 ⋮ 2a+3 => 2 ⋮ 2a+3 => 2a+3 ∈ Ư(2)

=> 2a+3 ∈ {1;2} => a ∈ ∅