\(ĐặtA=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(2A=1-\frac{1}{3^{2005}}\)
\(A=\frac{1-\frac{1}{3^{2005}}}{2}\)
Ủng hộ mk nha ^_-