Question

Tính:

B= 1+2+2^2+2^3+...+2^2008/1-2^2009

Answer 1

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

Đặt :

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

\(\Leftrightarrow2H=2+2^2+2^3+......+2^{2008}+2^{2009}\)

\(\Leftrightarrow2H-H=\left(2+2^2+.....+2^{2009}\right)-\left(1+2+2^2+....+2^{2008}\right)\)

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

\(\Leftrightarrow B=\frac{2^{2009}-1}{1-2^{2009}}=\frac{-\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)

Vậy...

Answer 2

Xét \(A=1+2+2^2+2^3+....+2^{2008}\)

\(\Leftrightarrow2A=2+2^2+2^3+...+2^{2009}\)

\(\Leftrightarrow2A-A=2^{2009}-1\)

Nên \(B=1\)