Question

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

Answer 1

Lời giải:

Xét tử số:
\(X=1+2+2^2+2^3+....+2^{2008}\)

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

Lấy vế sau trừ đi vế trước:

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

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

Do đó:

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

Answer 2

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

Ta xét tử số:

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

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

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

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

Thay vào B ta lại có:

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

Vậy B = -1