trả lời câu c nha

A=3+3^2 +3^+...+3^99+3^100

3A=3^2+3^3+...+3^100+3^101

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

Do đó 2A+3=3^101.Theo đề bài,2A+3=3^x

Vậy x=101

^ là mụ nha

3A=3^2+3^3+...+3^2007

=>3a-A=(3^2+3^3+...+3^2007)-(3^1+3^2+...+3^2006)

=>2A=3^2007-3^1=3^2007-3

=>2A+3=3^2007-3+3=3^2007=3^x

=>x=2007

 B=3+3^2+...+3^100.
3B=3.3+3^2.3+...+3^100.3
3B=3^2+3^3+...+3^101
3B-B=(3^2+3^3+...+3^101)-(3+3^2+...+3^100)
2B=3^101-3
Mà2B+3=3^n
Suy ra:3^101-3+3=3^n
3^n+3^101
Vậy n=101
Bài 1(b) làm tương tự,còn bài (a) thì bạn tự làm
 

mình giống nguyễn quỳnh nga

a,Ta có:3A=3²+3³+................+3²⁰¹¹

\(\Rightarrow3A-A=\left(3^2+3^3+.....+3^{2011}\right)-\left(3+3^2+.....+3^{2010}\right)\)

\(\Rightarrow2A=3^{2011}-3\)

\(\Rightarrow A=\frac{3^{2011}-3}{2}\)

b,Ta có:\(2A=3^{2011}-3\Rightarrow2A+3=3^{2011}\Rightarrow x=2011\)

3A - A = (3² + 3³ + 3⁴ + ... + 3²⁰⁰⁷) - (3 + 3² + 3³ + ... + 3²⁰⁰⁶)

2A = 3²⁰⁰⁷ - 3\(\Rightarrow\hept{\begin{cases}A=\frac{3^{2007}-3}{2}\\2A+3=3^{2007}\Rightarrow x=2007\end{cases}}\)

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

\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{2016}\right)\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2017}\)

\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+3^{2016}\right)\)

\(\Rightarrow2A=-3+3^{2017}\)

\(\Rightarrow A=\frac{3+3^{2017}}{2}\)

b) \(2A+3=-3+3-3^{2017}=3^{2017}=3^x\)

\(\Rightarrow x=2017\)

mình cũng ko biết câu này

\(A=3+3^2+3^3+...+3^{2006}\)

\(\Leftrightarrow3A=3\left(3+3^2+3^3+....+3^{2006}\right)\)

\(\Leftrightarrow3A=3^2+3^3+3^4+....+3^{2007}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2007}\right)-\left(3+3^2+3^3+...+3^{2006}\right)\)

\(\Leftrightarrow2A=3^{2007}-3\)

\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)

Ta có \(2A=3^{2007}-3\)

=> 2A+3=\(3^{2007}-3+3=3^{2007}\)

=> x=2007

A=3^1+3^2+3^3+....+3^2006

3A=3^2+3^3+...+3^2007

=>2A=3^2007-3

=>2A+3=3^x

3^2007-3+3=3^x

3^2007=3^x

=>x=2007

Vậy x=2007

Ta có : \(A=3+3^2+3^3+......+3^{2006}\)

=> \(3A=3^2+3^3+......+3^{2007}\)

=> \(3A-A=3^{2007}-3\)

=> \(2A=3^{2007}-3\)

=> \(A=\frac{3^{2007}-3}{2}\)

b) Ta có : \(2A=3^{2007}-3\) (theo ý a)

=> \(2A+3=3^{2007}\)

=> x = 2007

\(A=3+3^2+3^3+.........+3^{2006}\)

\(\Leftrightarrow3A=3^2+3^3+.........+3^{2007}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+.......+3^{2007}\right)-\left(3+3^2+.....+3^{2006}\right)\)

\(\Leftrightarrow2A=3^{2007}-3\)

\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)

\(\Leftrightarrow2A+3=3^{2007}\)

\(\Leftrightarrow3^x=3^{2007}\)

\(\Leftrightarrow x=2007\left(tm\right)\)

a)\(A=3+3^2+3^3+...+3^{2006}\)

=>\(3A=3\left(3+3^2+3^3+...+3^{2006}\right)=3^2+3^3+3^4+...+3^{2007}\)

=>\(3A-A=\left(3^2+3^3+3^4+...+3^{2007}\right)-\left(3+3^2+3^3+...+3^{2006}\right)\)

=>\(2A=3^{2007}-3\Rightarrow A=\frac{3^{2007}-3}{2}\)

b)\(2A+3=3^x\Rightarrow2.\frac{3^{2007}-3}{2}+3=3^x\Rightarrow3^{2007}=3^x\Rightarrow x=2007\)