Question

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

Tìm x để 2A+3=\(3^x\)

Trình bày lời giải đầy đủ nhé ,cảm ơn

Answer 1

ta có : \(A=3+3^2+3^3+...+3^{2008}\)

\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{2008}\right)\)

\(3A=3^2+3^3+3^4+...+3^{2009}\)

\(\Rightarrow3A-A=2A=\left(3^2+3^3+3^4+...+3^{2009}\right)-\left(3+3^2+3^3+...+3^{2008}\right)\)

\(2A=3^{2009}-3\) \(\Rightarrow2A+3=3^{2009}-3+3=3^{2009}=3^x\)

\(\Rightarrow x=2009\) vậy \(x=2009\)

Answer 2

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

\(\Leftrightarrow3A=3^2+3^3+3^4+....+3^{2009}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+....+3^{2009}\right)-\left(3+3^2+3^3+....+3^{2008}\right)\)

\(\Leftrightarrow2A=3^{2009}-3\)

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

Vậy x=3 để \(2A+3=3^x\)