\(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2B=2\cdot\left(2^{100}-2^{99}+2^{98}-...+2^2-2\right)\)
\(2B=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(2B+B=2^{101}-2^{100}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)
\(3B=2^{101}-2\)
\(B=\dfrac{2^{101}-2}{3}\)
\(B=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\\ =\left(2^{100}+2^{98}+...+2^2\right)-\left(2^{99}+2^{97}+...+2\right)\\ =\left(2^2+...+2^{98}+2^{100}\right)-\left(2+...+9^{97}+9^{99}\right)\\ =M+N\left(1\right)\)
Xét \(M=2^2+...+2^{98}+2^{100}\\ 4M=2^4+...+2^{100}+2^{102}\\ 4M-M=2^4+...+2^{100}+2^{102}-2^2-...-2^{98}-2^{100}\\ 3M=2^{102}-2^2\\ M=\dfrac{2^{102}-2^2}{3}\left(2\right)\)
Xét \(N=2+...+2^{97}+2^{99}\\ 4N=2^3+...+2^{99}+2^{101}\\ 4N-N=2^3+...+2^{99}+2^{101}-2-...-2^{97}-2^{99}\\ 3N=2^{101}-2\\ N=\dfrac{2^{101}-2}{3}\left(3\right)\)
Từ `(1);(2)` và `(3)` suy ra
\(B=\dfrac{2^{102}-2^2}{3}-\dfrac{2^{101}-2}{3}\\ =\dfrac{2^{102}-2^2-2^{101}+2}{3}=\dfrac{2^{101}\left(2-1\right)-2}{3}\\ =\dfrac{2^{101}-2}{3}\)
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