đặt A=1 +2 + \(2^2\)+ \(2^3\)+...+\(2^{2008}\)
2A= 2 +\(2^2\)+\(2^3\)+...+\(2^{2008}\)+\(2^{2009}\)
ta có 2A-A= \(2^{2009}\)-1
A=\(2^{2009}\)-1
vậy B = \(\frac{1+2+2^2+.....+\text{2^{2008}}}{1-2^{2009}}\)=\(\frac{2^{2009}-1}{1-2^{2009}}\)=-1
vậy B=-1
\(1+2+2^2+2^3+...+2^{2008}\)
\(2B=2+2^2+2^3+...+2^{2009}\)
\(2B-B=B=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)\)
\(2^{2009}-1\)
\(\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
\(\frac{2^{2009}-1}{1-2^{2009}}\)
\(-\left(\frac{1-2^{2009}}{1-2^{2009}}\right)=1\)
mà mk cx nhớ là mk có hỏi ai đó ùi mà
