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