Ta có :

A=3+3²+...+3²⁰¹⁵

=> 3A-A=3²+3³+...+3²⁰¹⁶- (3+3²+...+3²⁰¹⁵)

=>2A=3²⁰¹⁶-3

lại có: 2A+3=3ⁿ

=>3²⁰¹⁶-3+3=3ⁿ

=>3²⁰¹⁶=3ⁿ

=>n=2016

Vậy n=2016

A=3+3²+3³+...+3¹⁰⁰

=>3A=3²+3³+...+3¹⁰⁰

=>3A=A=3¹⁰¹-3

=>2A=3¹⁰¹-3

=>2A+3=3¹⁰¹

=>n=101

Vậy số tự nhiên n bằng 101

Có A = 3 + 3\(^2\) + ....... + 3 \(^{100}\) 

 3A   = 3\(^2\) + 3 \(^3\) + ...... + 3\(^{101}\) 

Lấy 3A - A 

\(\Rightarrow\) 2A = 3\(^{101}\) - 3

\(\Rightarrow\) 2A + 3 = 3\(^{101}\)

Mà theo bài ra, 2A + 3 = 3\(^n\)

\(\Rightarrow\) n = 101 

A=3+3²+3³+.....+3¹⁰⁰

3A=3.(3+3²+3³+....+3¹⁰⁰)

3A=3²+3³+3⁴+....+3¹⁰¹

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

2A=3¹⁰¹-3

2A+3=3¹⁰¹-3+3

2A+3=3¹⁰¹

3ⁿ=3¹⁰¹

=>n\(\in\)(101)

Chúc bn học tốt

Ta có: 3A=3²+3³+...+3¹⁰¹

3A-A=2A=(3²+3³+...+3¹⁰¹)-(3+3²+...+3¹⁰⁰)

2A=3¹⁰¹-3

A=\(\frac{3^{101}-3}{2}\)

=>2A+3=2.\(\frac{3^{101}-3}{2}\)+3

            =(3¹⁰¹-3)+3

           =3¹⁰¹

Mà 2A+3=3ⁿ

=>3¹⁰¹=3ⁿ

=>n=101

A=3+3²+3³+...+3¹⁰⁰

2A=(3+3²+3³+...+3¹⁰⁰)x2

2A=3²+3³+3⁴...+3¹⁰¹

2A-A=3¹⁰¹-3

mà 3ⁿ=2A+3=3¹⁰¹-3+3=3¹⁰¹

suy ra n=101

Ta có : A = 3 + 3² + 3³ + ... + 3¹⁰⁰

3A = 3²+3³+3⁴+...+3¹⁰¹

Vậy 2A = 3¹⁰¹ - 3 

Vậy 2A + 3 = 3¹⁰¹

=> x = 101

Ta có \(A=3+3^2+3^3+...+3^{100}\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{101}\)

\(\Rightarrow3A-A=3^{101}-3\)

\(2A=3^{101}-3\)

Ta có \(2A+3=3^n\)

hay \(3^{101}-3+3=3^n\)

\(3^{101}=3^n\)

\(n=101\)

A=3+3²+3³+.....+3¹⁰⁰

3a=3.(3+3²+3³+....+3¹⁰⁰)

3A=3²+3³+3⁴+....+3¹⁰¹

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

2A=3¹⁰¹-3

2A+3=3¹⁰¹-3+3

2A+3=3¹⁰¹

3ⁿ=3¹⁰¹

=>n\(\in\)(101)

Chúc bn học tốt

A=\(3+3^2+3^3+...+3^{100}\)

3A=\(3^2+3^3+3^4+...+3^{101}\)

3A - A=\(3^2+3^3+3^4+...+3^{101}-3-3^2-3^3-...-3^{100}\)

 2A = \(3^{101}-3\)

 =>\(2A+3=3^n\)

 =>\(3^{101}-3+3=3^n\)

 =>3\(^{101}=3^n\)

=>n=101

có A=3+3^2+3^3+..+3^100

3A=3.3+3^2.3+3^3.3+..+3^100.3

3A=3^2+3^3+3^4+..+3^101
⇒2A=(3^2+3^3+3^4+..+3^101)-(3+3^2+3^3+..+3^100)

2A=3^101-3

LẤY 3^101-3+3=3^n

3^101=3^n

⇒n=101

Ta có A = 3 + 3^2 + 3^3 + ... +3^{100}A=3+32+33+...+3100 (1)

3A = 3^2 + 3^3 + ... +3^{100} + 3^{101}3A=32+33+...+3100+3101 (2)

Lấy (2) trừ (1) được 2A = 3^{101} - 32A=3101−3.

Do đó, 2A + 3 = 3^{101}2A+3=3101

Mà theo đề bài 2A + 3 = 3^n2A+3=3n.

Vậy $n=101$.

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

3A=32+33+...+3100+31013A=32+33+...+3100+3101 (2)

Lấy (2) trừ (1) được 2A=3101−32A=3101−3.

Do đó, 2A+3=31012A+3=3101

Mà theo đề bài 2A+3=3n2A+3=3n.

Vậy n=101n=101.

=>3A=3²+3²+…+3¹⁰¹

=>3A-A=3²+3³+…+3¹⁰¹-3-3²-…-3¹⁰⁰

=>2A=3¹⁰¹-3

=>2A+3=3¹⁰¹=3^N

=>N=101

Vậy N=101

3A = \(3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{100}+3^{101}\right)\)- \(\left(3+3^2+3^3+..+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\Rightarrow2A+3=3^{101}\)
Vậy n = 101

3A=3²+3³+3⁴+3⁵+......+3¹⁰¹

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

2A=3¹⁰¹-3

2A+3=3¹⁰¹-3+3

2A+3=3¹⁰¹ 

Mà theo đề bài thì 2A+3=3ⁿ suy ra n=101

A = 3¹⁰⁰ + 3

\(A=3+3^2+3^3+...+3^{100}\)

\(3A=3^2+3^3+...+3^{100}+3^{101}\)

\(3A-A=3^2+3^3+...+3^{100}+3^{101}-\left(3+3^2+3^3+...+3^{100}\right)\)

\(2A=3^{101}-3\)

\(2A+3=3n\)

\(\Rightarrow3^{101}-3+3=3n\)

\(\Rightarrow3^{101}=3n\)

\(\Rightarrow n=3^{100}\)

Ta có A= 3+3^2+3^3+...+3^100

        3A= 3^2+3^3+3^4+...+3^101

   3A-A=(3^2+3^3+3^4+...+3^101)-(3+3^2+3^3+...+3^100)

     2A= 3^101 - 3

Ta lại có 2A+3=3^101-3+3

                    = 3^101

=> 3n=3^101

=> n= 3^101:3

=> n= 3^100

Vậy n= 3^100

A=3+3^2+3^3+...+3^100

=>3A=3^2+3^3+3^4+...+3^101

=>3A-A=2A=3^101-3

mà 2A+3=3^n

=>3^101-3+3=3^n

=>3^n=3^101

=>n=101