Question

Cho H=\(2^{2010}-2^{2009}-2^{2008}-...-2-1\) .Tính \(2010^H\)

Answer 1

Ta có:

\(=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)

Đặt \(A=2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow2A=2^{20010}+2^{2009}+...+2^2+2\)

\(\Rightarrow2A-A=\left(2^{20010}+2^{2009}+...+2^2+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)\(\Rightarrow A=\left(2^{2010}-1\right)+\left(2^{2009}-2^{2009}\right)+\left(2^{2008}-2^{2008}\right)+...+\left(2-2\right)\)\(\Rightarrow A=2001-1\)

\(\Rightarrow H=2^{2010}-\left(2^{2010}-1\right)\)

\(\Rightarrow H=2^{2010}-2^{2010}+1=1\)

Thay \(H=1\) vào biểu thức \(2010^H\)

\(\Rightarrow2010^H=2010^1=1\)

Vậy \(2010^H=1\)