Question

tính:
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

Answer 1

Đặt \(A=2^{2009}+2^{2008}+...+2^1+2^0\)

Ta có : \(2A=2^{2010}+2^{2009}+...+2^2+2^1\)

\(\Rightarrow2A-A=2^{2010}-2^0\Rightarrow A=2^{2010}-1\)

Do đó : \(M=2^{2010}-A=2^{2010}-\left[2^{2010}-1\right]=1\)

Answer 2

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

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

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

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

\(2\left(2^{2010}-M\right)-M=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)

\(2^{2010}-M=2^{2010}+2^{2009}+...+4+2-2^{2009}-2^{2008}-...-2-1\)

\(2^{2010}-M=2^{2010}-1\)

=> M = 1