\(B=\dfrac{1+2+2^2+...+2^{2008}}{1-2^{2009}}\)
\(2B=\dfrac{2+2^2+2^3+...+2^{2009}}{1-2^{2009}}\)
\(B-2B=\)\(\dfrac{1+2+2^2+...+2^{2008}}{1-2^{2009}}\)\(-\dfrac{2+2^2+2^3+...+2^{2009}}{1-2^{2009}}\)
\(-B=\dfrac{1-2^{2009}}{1-2^{2009}}\)
B=-1
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 \(\dfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
B= \(\dfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
giúp mình nha !!!!!!!!!!
Tính tổng S= \(\dfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
Giúp mình với mình đang cần gấp!!!
a, \(\dfrac{5}{2}-3\left(\dfrac{1}{3}-x\right)=\dfrac{1}{4}-7x\)
b, \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2008}\right).x=\dfrac{2009}{1}+\dfrac{2010}{2}+...+\dfrac{5016}{2008}-2008\)
c, \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2001}{2003}\)
GIÚP VỚI , MIK CẦN GẤP
Tính A= 2009/2+2008/(2^2)+2007/(2^3)+...+3/(2^2007)+2/(2^2008)+1/(2^2009)
Tính nhanh:
a) A = 2\(^{2010}\) - 2\(^{2009}\) - 2\(^{2008}\) - 2\(^{2007}\) - ... - 2 - 1
b) B = 20 . 8 - 33 . 64 + 17 . 8 + 9 . 16 . 8 - 11 . 128
c) C = ( \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + ... + \(\dfrac{1}{2009.2010.2011}\) ) . \(\dfrac{2010.2011}{1010.527}\)
so sánh 2 phân số : \(A=\frac{2008^{2009}+2}{2008^{2009}-1};B=\frac{2008^{2009}}{2008^{2009}-3}\)
Bài 5:1/2+1/3+...+1/2008).x=2009/1+2010/2+2011/3+5016/2008-2008
