Question

[Vũ Thị Hải Yến]

Tính B

Answer 1

[Mai Linh]

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

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

ta có 2A-A= \(2^{2009}\)-1

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

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

vậy B=-1

Answer 2

[Đặng Thị Cẩm Tú]

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

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

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

\(2^{2009}-1\)

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

\(\frac{2^{2009}-1}{1-2^{2009}}\)

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

mà mk cx nhớ là  mk có hỏi ai đó ùi mà