Question

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

Tìm x để 2A+3=3^x

Answer 1

Ta có:

\(A=3^1+3^2+3^3+...+3^{2006}\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2007}\)

\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2007}\right)-\left(3^1+3^2+3^3+...+3^{2006}\right)\)

\(\Rightarrow2A=3^{2007}-3\)

Thay vào biểu thức:

\(2A+3=3^x\)

\(\Rightarrow3^{2007}-3+3=3^x\)

\(\Rightarrow3^{2007}=3^x\)

\(\Rightarrow x=2007\)

Answer 2

Ta có

\(A=3^1+3^2+3^3+...+3^{2006}\)

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

⇒\(3A-A\)=( \(3^2+3^3+3^4+...+3^{2007} \) )−( \(3^1+3^2+3^3+...+3^{2006}\))

⇒\(2A=3^{2007}-3\)

⇒\(2A+3=3^{2007}\)

Mà \(2A+3=3^x\)

=> \(x=2007\)

Vậy..