Question

[Nguyễn Trần Nguyệt Kỳ]

2¹⁰⁰- (2⁹⁹+2⁹⁸+2⁹⁷+...+2+1)

Các bạn giúp mik vs, mik đang cần gấp

Answer 1

[Tạ Đức Hoàng Anh]

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 \(2^{100}-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)=1\)