\(S=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
3 tháng 3 2015
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
26 tháng 5 2018
1.
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
cứ làm như vậy ta được :
\(=1+1=2\)
26 tháng 5 2018
2. Ta có :
\(\frac{2008+2009}{2009+2010}=\frac{2008}{2009+2010}+\frac{2009}{2009+2010}\)
vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)
GG
0
TQ
1
TN
17 tháng 4 2016
đặt tử =A,ta có:
tử=2A=2(1+2.2+2.22+...+2.22008)
=2.1+2.2+2.22+...+2.22008
=2+22+23+...+22009
2A-A=(2+22+23+...+22009)-(1+2+22+...+22008)
A=22009-1
thay A vào tử của S ta được:\(S=\frac{2^{2009}-1}{1-2^{2009}}=-1\)
BB
31 tháng 8 2020
Ta có: \(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...............+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+........+\left(1+\frac{1}{2008}\right)+1\)
\(B=\frac{2009}{2}+\frac{2009}{3}+..............+\frac{2009}{2008}+\frac{2009}{2009}\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{2009}\right)\)
Khi đó: \(\text{}\text{}\text{}\frac{A}{B}=\frac{1}{2009}\)
Chuc bạn học tốt!!
30 tháng 8 2020
Ta có: \(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=2008+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)\)
\(=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}\)
\(=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}\right)\)
Ta có: \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}}{2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}\right)}\)
hay \(\frac{A}{B}=\frac{1}{2009}\)
NL
0
Coi: \(C=1+2+2^2+2^3+...+2^{2008}\)
\(\Rightarrow2C=2.\left(1+2+2^2+2^3+...+2^{2008}\right)=2+2+2^2+...+2^{2007}\)
\(\Rightarrow C=2C-C=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)=2-2^{2008}\)
\(\Rightarrow S=\frac{2-2^{2008}}{1-2^{2009}}\)
Coi: $C=1+2+2^2+2^3+...+2^{2008}$C=1+2+22+23+...+22008
$\Rightarrow2C=2.\left(1+2+2^2+2^3+...+2^{2008}\right)=2+2+2^2+...+2^{2007}$
Coi: $C=1+2+2^2+2^3+...+2^{2008}$C=1+2+22+23+...+22008
$\Rightarrow2C=2.\left(1+2+2^2+2^3+...+2^{2008}\right)=2+2+2^2+...+2^{2007}$
Coi: $C=1+2+2^2+2^3+...+2^{2008}$C=1+2+22+23+...+22008
$\Rightarrow2C=2.\left(1+2+2^2+2^3+...+2^{2008}\right)=2+2+2^2+...+2^{2007}$
Coi: $C=1+2+2^2+2^3+...+2^{2008}$C=1+2+22+23+...+22008
$\Rightarrow2C=2.\left(1+2+2^2+2^3+...+2^{2008}\right)=2+2+2^2+...+2^{2007}$