Question

2. tính

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

Answer 1

Đặt \(A=1+2+2^2+...+2^{2008}\)

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

\(2A-A=2+2^2+2^3+...+2^{2009}-\left(1+2+2^2+...+2^{2008}\right)\)

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

mà \(B=\dfrac{A}{1-2^{2009}}=-1\)

Answer 2

Giả sử ta Đặt A= \(1+2+2^2+2^3+...+2^{2008}\)

Ta có 2A = \(2+2^2+2^3+...+2^{2009}\)

Lấy 2A - A = \(2+2^2+2^3+...+2^{2009}-\left(1+2+2^2+...+2^{2008}\right)\)

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

Ta lại có \(B=\dfrac{A}{1-2^{2009}}\)

Thay A vào B ta có \(B=\dfrac{2^{2009}-1}{1-2^{2009}}=\dfrac{-\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)