Question

Tính

\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)

Answer 1

Đặt A \(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^{2005}}\)

\(\Rightarrow A=\left(1-\frac{1}{3^{2005}}\right):2\)

Answer 2

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(\Leftrightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

\(\Leftrightarrow3A-A=1-\frac{1}{3^{2005}}\)

\(\Leftrightarrow A=\frac{1-\frac{1}{3^{2005}}}{2}\)

Answer 3

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(3A=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

\(3A-A=1-\frac{1}{3^{2005}}\)

\(A=\frac{1-\frac{1}{3^{2005}}}{2}\)

Answer 4

a=1+3+3^2+...+3^100
=> 3a=3+3^2+...+3^101
=>2a=3^101-1
=>a=(3^101-1):2