Đặt \(A=1+2+...+2^{97}+2^{98}+2^{99}\)\(\Rightarrow\)\(2^{100}-A=2^{100}-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)\)
Ta có: \(2A=2+2^2...+2^{98}+2^{99}+2^{100}\)
Lấy \(2A-A\)theo vế, ta có:
\(2A-A=\left(2+2^2...+2^{98}+2^{99}+2^{100}\right)-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)\)
\(\Leftrightarrow2A-A=2+2^2...+2^{98}+2^{99}+2^{100}-1-2-...-2^{97}-2^{98}-2^{99}\)
\(\Leftrightarrow A=2^{100}-1\)
\(\Rightarrow2^{100}-A=2^{100}-2^{100}+1=1\)
Vậy \(2^{100}-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)=1\)
