Đặt \(A=1-3+3^2-3^3+...-3^{99}+3^{100}\)
\(\Rightarrow3A=3-3^2+3^3-...-3^{100}+3^{101}\)
\(\Rightarrow3A+A=3-3^2+3^3-...-3^{100}+3^{101}+1-3+3^2-3^3+...-3^{99}+3^{100}\)
\(\Rightarrow4A=1+3^{101}\)
\(\Rightarrow A=\dfrac{1+3^{101}}{4}\)
Ta có: \(A=1-3+3^2-3^3+...-3^{99}+3^{100}\)
\(\Leftrightarrow3\cdot A=3-3^2+3^3-3^4+...-3^{100}+3^{101}\)
\(\Leftrightarrow4\cdot A=3^{101}+1\)
hay \(A=\dfrac{3^{101}+1}{4}\)