Giải:
Ta có:
\(A=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\left(1\right)\)
\(\Rightarrow3A=3\left(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\right)\)
\(=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\left(2\right)\)
Lấy \(\left(2\right)+\left(1\right)\Leftrightarrow4A=3^{101}+1\)
\(\Rightarrow A=\dfrac{3^{101}+1}{4}\)