\(A=3^{100}-3^{99}+3^{98}-3^{97}+.........+3^2-3+1\)
\(\Leftrightarrow3A=3^{101}-3^{100}+3^{99}-3^{98}+......+3^3-3^2+3\)
\(\Leftrightarrow3A+A=\left(3^{101}-3^{100}+3^{99}-3^{98}+........+3^3-3^2+3\right)+\left(3^{100}-3^{99}+3^{98}-3^{97}+.......+3^2-3+1\right)\)
\(\Leftrightarrow4A=3^{101}+1\)
\(\Leftrightarrow A=\dfrac{3^{101}+1}{4}\)