Question

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

Answer 1

Gọi \(1+2+2^2+2^3+...+2^{2008}\) là D.

Ta có:

\(D=1+2+2^2+2^3+...+2^{2008}\)

\(2D=2+2^2+2^3+2^4...+2^{2009}\)

\(2D-D=\left(2+2^2+2^3+2^4...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)\)\(D=2^{2009}-1\)

\(B=\dfrac{2^{2009}-1}{1-2^{2009}}\\ =\dfrac{\left(-1\right)\cdot\left(1-2^{2009}\right)}{1-2^{2009}}\\ =-1\)