\(=\dfrac{2\left(1+2+2^2+...+2^{2008}\right)-\left(1+2+2^2+...+2^{2008}\right)}{1-2^{2009}}\)
\(=\dfrac{\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+...+2^{2008}\right)}{1-2^{2009}}\)
\(=\dfrac{2^{2009}-1}{1-2^{2009}}=-1\)
\(=\dfrac{2\left(1+2+2^2+...+2^{2008}\right)-\left(1+2+2^2+...+2^{2008}\right)}{1-2^{2009}}\)
\(=\dfrac{\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+...+2^{2008}\right)}{1-2^{2009}}\)
\(=\dfrac{2^{2009}-1}{1-2^{2009}}=-1\)
Tính tổng S= \(\dfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
Giúp mình với mình đang cần gấp!!!
tính tổng S=1+2+2^2+2^3+.....+2^2008 / 1-2^2009
Tính tổng S=1+2+2^2+2^3+...+2^2008/1-2^2009
1. \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
2. So sánh: \(\dfrac{2008}{2009}+\dfrac{2009}{2010}\) và \(\dfrac{2008+2009}{2009+2010}\)
Tính tổng sau
A=1-2+3-4+5-....-2008+2009
B=1+2-3-4+5+6-7-...-2007-2008+2009+2010
Tính:
B=\(\dfrac{1+2+2^2+2^3+...2^{2008}}{1-2^{2009}}\)
B= \(\dfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
giúp mình nha !!!!!!!!!!
Tính A= 2009/2+2008/(2^2)+2007/(2^3)+...+3/(2^2007)+2/(2^2008)+1/(2^2009)
Tính tổng C= 1+2+2^2+....+2^2008/ 1- 2^2009