Đặt A=1+2+22+23+...+22008
=>2A=2+22+23+24+...+22009
=>2A-A=A=(2+22+23+24+...+22009)-(1+2+22+23+...+22008)
=22009-1
Suy ra:\(\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}=\frac{2^{2009}-1}{1-2^{2009}}=\frac{-\left(1-2^{2009}\right)}{1-2^{2009}}=-1\)
Tính: B=\(\dfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
1) Cho : A= 1.2.3.4.........2008(1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.........+\frac{1}{2008}\))
CMR : A chia hết cho 2009
tính
a)B = 1+2+2^2+2^3+...+2^2008/1-2^2009
Tính:
B= \(1+2+2^2+2^3+...+2^{2008}\)/ 1-22009
B= 1+2 + 2 mũ 2+ 2 mũ 3+.....+2 mũ 2008/1- 2 mũ 2009
( cái mk viết / có nghĩa là phần)
Chứng minh rằng
A = 1.2.3.....2007.2008.\(\left(1+\frac{1}{2}+...+\frac{1}{2007}+\frac{1}{2008}\right)\) chia hết cho 2009
Chứng minh rằng hiệu sau là 1 số nguyên
\(\frac{10^{2008}+2}{3}\)-\(\frac{10^{2009}+17}{9}\)
\(\frac{x-1}{2009}\)+\(\frac{x-2}{2008}\) = \(\frac{x-3}{2007}\)+\(\frac{x-4}{2006}\)
1+2-3-4+5-6-7-8+9+10-...+2006-2007-2008+2009
