Question

Cho A=1+3+\(3^2\)+\(3^3\)+...\(3^{10}\)

Answer 1

\(A=1+3+3^2+...+3^{10}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{11}\)

\(\Rightarrow3A-A=2A=3^{11}-1=177146\)

\(\Rightarrow A=\frac{177146}{2}=88573\)

Answer 2

\(A=1+3+3^2+...+3^{10}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{11}\)

\(\Rightarrow3A-A=3^{11}-1\)

\(\Leftrightarrow2A=3^{11}-1\)

\(\Rightarrow A=\frac{3^{11}-1}{2}\)