Câu hỏi của Thuy Khuat

Question

Cho H=$2^{2010}-2^{2009}-2^{2008}-...-2-1$ .Tính $2010^H$

Lê Minh Tuấn · Accepted Answer

Ta có: $H=2^{2010}-2^{2009}-2^{2008}-...-2-1$

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

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 $2010^H=1$Câu trả lời: Ta có: $H=2^{2010}-2^{2009}-2^{2008}-...-2-1$