=>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

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

3/4 +3 =

loai j vay ban

3A = 3^2 + 3^3 + ..... + 3^2010

3A - A = 2A = (3^2 + 3^3 + ..... + 3^2010) - (3 + 3^2 + 3^3 + ..... + 3^2009)

                  = 3^2010 - 3

=> 2A + 3 = 3^2010 - 3 + 3 = 3^2010 = 3^n

suy ra n = 2010

chắc đúng r đấy bn :D ^^

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

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

\(\Rightarrow\)2A + 3 = 3^101  \(-\)3  +  3  =  3^101

\(\Rightarrow\)3^N  =  3^101

\(\Rightarrow\)N = 101

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

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

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

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

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

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 $A=3+3^2+3^3+...+3^{100}$ (1)

$3A=3^2+3^3+...+3^{100}+3^{101}$ (2)

Lấy (2) trừ (1) được $2A=3^{101}-3$.

Do đó, $2A+3=3^{101}$

Mà theo đề bài $2A+3=3^n$.

Vậy $n=101$.