M = \(2^{100-}2^{99}+2^{98}-...+2^2-2\)
\(2M=2^{101}-2^{100}+2^{99}-2^{98}+2^{97}+...+2^3-2^2\)
\(2M+M=2^{101}-2\)
\(M=\frac{2^{101}-2}{3}\)
N=\(3^{100}-3^{^{ }99}+3^{98}-3^{97}+...+3^2-3+1\)
\(3N=3^{101}-3^{100}+3^{99}-3^{98}+3^{97}+...+3^3-3^2+3\)
3N+N= 4N = \(3^{101}+1\)
N=\(\frac{3^{101}+1}{4}\)