Question

bài 3: 

bài 4: 

Answer 1

Bài 3:

\(A=3+3^2+3^3+...+3^{100}\)

\(\Rightarrow3A=3^2+3^3+...+3^{101}\)

\(\Rightarrow2A+3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}=3^{101}-3\)

\(\Rightarrow2A+3=3^{101}=3^n\)

\(\Rightarrow n=101\)

Answer 2

\(A=3+3^2+...+3^{100}\Leftrightarrow3A=3^2+3^3+...+3^{101}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\\ \Leftrightarrow2A=3^{101}-3\\ \Leftrightarrow2A+3=3^{101}\\ \Rightarrow n=101\)