Question

Cho A = 3 +3^2 + 3^3 + .....+ 3^2008. tim x biet 2a+3=3^x

 

Answer 1

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

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

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

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

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

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

Vì \(2A+3=3^x\)hay \(3^{2009}=3^x\)

 \(\Rightarrow x=2009\)

Answer 2

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