\(B=\dfrac{1+2+2^2+2^3+.....+2^{2008}}{1-2^{2009}}\)
Đặt \(S=1+2+2^2+2^3+....+2^{2008}\)
\(2S=2\left(1+2+2^2+2^3+....+2^{2008}\right)\)
\(2S=2+2^2+2^3+2^4+.....+2^{2009}\)
\(2S-S=\left(2+2^2+2^3+2^4+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)\)\(S=2^{2009}-1\)
Thay S vào B ta có:
\(B=\dfrac{1-2^{2009}}{2^{2009}-1}=-1\)
\(B=\dfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}.\)
Đặt phần tử của \(B\) là \(C\Rightarrow B=\dfrac{C}{1-2^{2009}}.\)
Ta có:
\(C=1+2+2^2+2^3+...+2^{2008}.\)
\(2C=2\left(1+2+2^2+2^3+...+2^{2008}\right).\)
\(2C=2+2^2+2^3+2^4+...+2^{2009}.\)
\(2C-C=\left(2+2^2+2^3+2^4+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right).\)
\(C=\left(2-2\right)+\left(2^2-2^2\right)+\left(2^3+2^3\right)+...+\left(2^{2008}-2^{2008}\right)+\left(2^{2009}-1\right).\)
\(C=0+0+0+...+0+\left(2^{2009}-1\right).\)
\(C=2^{2009}-1.\)
Thay \(C\) vào \(B.\)
\(\Rightarrow B=\dfrac{C}{1-2^{2009}}=\dfrac{2^{2009}-1}{1-2^{2009}}=-1.\)
\(\Rightarrow B=-1.\)
Vậy.....
~ Học tốt!!! ~
