Question

B=\(\frac{1+2+2^2+2^3+....+2^{2008}}{1-2^{2009}}\)

Answer 1

Đặt \(A=1+2+2^2+2^3+...+2^{2008}\)

\(2A=2\left(1+2+2^2+2^3+...+2^{2008}\right)\)

\(2A=2+2^2+2^3+...+2^{2009}\)

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

\(A=2^{2009}-1=-\left(1-2^{2009}\right)\)

Khi đó \(B=\dfrac{A}{1-2^{2009}}=\dfrac{-\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)

P/s: 1 .tìm trước khi hỏi

2. Nhớ để câu hỏi đúng box tái phạm thì đừng trách