Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(\Rightarrow5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(\Rightarrow5A-A=\left(1-\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\right)\)
\(4A=1-\frac{1}{5^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{5^{100}}}{4}\)
Vậy \(A=\frac{1-\frac{1}{5^{100}}}{4}\)