Question

tính \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2008}}\)

Answer 1

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2008}}\)

\(3.A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\)

\(3.A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2008}}\right)\)

\(\Rightarrow2.A=1-\frac{1}{3^{2008}}\)

\(\Rightarrow A=\left(1-\frac{1}{3^{2008}}\right):2\)

Answer 2

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2008}}\)

\(3A=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2008}}\right)\)

\(3A=1+\frac{1}{3}+...+\frac{1}{3^{2007}}\)

\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2008}}\right)\)

\(2A=\frac{1}{3^{2007}}-\frac{1}{3}\)

\(A=\frac{\frac{1}{3^{2007}}-\frac{1}{3}}{2}\)