Ta có: \(A=2^{100}-2^{99}-2^{98}-...-2\)
\(\Rightarrow A=2^{100}-\left(2^{99}+2^{98}+...+2\right)\)
Đặt \(B=2^{99}+2^{98}+...+2\)
\(\Rightarrow2B=2^{100}+2^{99}+...+2^2\)
\(\Rightarrow2B-B=\left(2^{100}+2^{99}+...+2^2\right)-\left(2^{99}+2^{98}+...+2\right)\)
\(\Rightarrow B=2^{100}-2\)
\(\Rightarrow A=2^{100}-\left(2^{100}-2\right)\)
\(\Rightarrow A=2^{100}-2^{100}+2\)
\(\Rightarrow A=2\)
Vậy A= 2
A = 2^100 - 2^99 - 2^98 - ... - 2^2 - 2
A = 2^100 - (2^99 + 2^98 + ... + 2^2 + 2)
Đặt B = 2^99 + 2^98 + ... + 2^2 + 2
2B = 2^100 + 2^99 + ... + 2^3 + 2^2
2B - B = 2^100 - 2 = B
A = 2^100 - B = 2^100 - (2^100 - 2)
A = 2^100 - 2^100 + 2 = 2
