Gọi \(S=1+2+2^2+2^3+...+2^{2008}\)
\(S=1+2+2^2+2^3+...+2^{2008}\\ 2S=2+2^2+2^3+2^4+...+2^{2009}\\ 2S-S=\left(2+2^2+2^3+2^4+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)\\ S=2^{2009}-1\)
\(\dfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\\ =\dfrac{2^{2009}-1}{\left(-1\right)\cdot\left(2^{2009}-1\right)}\\ =\dfrac{1}{-1}\\ =-1\)
Đặt A=\(1+2+2^2+...+2^{2008}\)
=>2A=2.(\(1+2+2^2+...+2^{2008}\))
=>2A=\(2+2^2+2^3+...+2^{2009}\)
=>2A-A=(\(2+2^2+2^3+...+2^{2009}\))-(\(1+2+2^2+...+2^{2008}\))
=>A=\(2^{2009}-1\)
=>A=(-1).\(\left(-2\right)^{2009}\)+(-1).1
=>A=(-1).\(\left[\left(-2\right)^{2009}+1\right]\)
=>A=(-1).\(\left(1-2^{2009}\right)\)
=>\(1+2+2^2+...+2^{2008}\)/1-\(2^{2009}\)
=\(\dfrac{\left(-1\right).\left(1+2^{2009}\right)}{1+2^{2009}}\)=-1
