Giải
\(2.B=\frac{2.\left(1+2+2^3+2^4+...+2^{2008}\right)}{1-2^{2009}}\)
2B= \(\frac{2^1+2^2+...+2^{2009}}{1-2^{2009}}\)
2B - B = B = \(\frac{2+2^2+...+2^{2009}}{1-2^{2009}}-\frac{1+2+2^2+...2^{2008}}{1-2^{2009}}\)
B = \(\frac{2^{2009}-1}{1-2^{2008}}\)
Xét tử số: A=1+2+22 +...+ 22008
=> 2A = 2+22 + 23 +...+ 22009
=> 2A-A = 22009 - 1 =A
=> B=\(\frac{2^{2009}-1}{1-2^{2009}}\) = \(\frac{\left(-1\right).\left(1-2^{2009}\right)}{1-2^{2009}}\) = -1
đặt tử là A ta có:
2A=2(1 + 2 + 22 + 23 +... + 22008 )
2A=2+22+23+...+22009
2A-A=(2+22+23+...+22009)-(1 + 2 + 22 + 23 +... + 22008)
A=22009-1
thay A vào tử ta đc \(B=\frac{2^{2009}-1}{1-2^{2009}}=-1\)
đặt tử là A ta có:
2A=2(1 + 2 + 22 + 23 +... + 22008 )
2A=2+22+23+...+22009
2A-A=(2+22+23+...+22009)-(1 + 2 + 22 + 23 +... + 22008)
A=22009-1
thay A vào tử ta đc $B=\frac{2^{2009}-1}{1-2^{2009}}=-1$