Tính \(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2008}}\)
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3 tháng 3 2015
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
GT
2
8 tháng 11 2017
5G= 1+1/5+1/5^2+.....+1/5^2007
4G=5G-G=(1+1/5+1/5^2+....+1/5^2007)-(1/5+1/5^2+1/5^3+....+1/5^2008)
= 1 - 1/5^2008
=>G=(1-1/5^2008)/4
8 tháng 11 2017
\(G=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2008}}\)(1)
\(\Rightarrow5G=1+\frac{1}{5}+...+\frac{1}{5^{2007}}\)(2)
Lấy (2) trừ đi (1) ta có :
\(4G=1-\frac{1}{5^{2008}}\)
\(\Rightarrow G=\frac{\left(1-\frac{1}{5^{2008}}\right)}{4}\)
HT
29 tháng 8 2015
Xét tử ta có:
\(2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{1}{2008}\)
= \(1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{1}{2008}\right)\)
= \(\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}\)
= \(2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\)
=> A = \(\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}}\)
=> A = 2009
29 tháng 8 2015
A=\(\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...........+\left(1+\frac{2}{2008}\right)+\left(1+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2008}+\frac{1}{2009}}\)=\(\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+....+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\frac{ }{ }\)
=\(\frac{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2008}+\frac{1}{2009}}\frac{ }{ }\)
=2009
Vay A=2009
1 tháng 5 2016
2008 + 2007/2 + 2006/3 + 2005/4 + ... + 2/2007 + 1/2008
2009-1/1 + 2009-2/2 + 2009-3/3 + 2009-4/4 + ... + 2009-2007/2007 + 2009-2008/2008
2009 - 1 + 2009/2 - 1 + 2009/3 - 1 + 2009/4 - 1 + ... + 2009/2007 - 1 + 2009/2008 - 1
2009 + 2009.(1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 ) - ( 1 + 1 + 1 + 1 + ... + 1 + 1 )
2009 + 2009.( 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 ) - 2008
1 + 2009.( 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 )
2009.( 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009 )
=> giá trị của biểu thức trên là 2009
NL
0
18 tháng 4 2017
2008=1+1+1+...+1 có 2008 số 1
1+(1+2007/2)+(1+2006/3)+...+(1+1/2008)=2009/2009+2009/2+2009/3+...+2009/2008
=2009*(1/2009+1/2+1/3+...+1/2008)=2009*(1/2+1/3+...+1/2009)
ta có 2008+2007/2+...+1/2008
1/2+1/3+..............+1/2009
=2009
TG
23 tháng 10 2016
Gọi a là tử số, b là mẫu số của phân số A
a = \(\frac{2008}{1}\)+ \(\frac{2007}{2}\)+ \(\frac{2006}{3}\)+ ... + \(\frac{1}{2008}\)
Dãy số a có (2008 - 1) : 1 + 1 = 2008 số. Và a = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2)
b = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{2009}\)
Dãy số b có (2009 - 2) : 1 + 1 = 2008 số. Và b = (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)
A = [ ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) x (2008 : 2)] : [ (\(\frac{1}{2}\)+ \(\frac{1}{2009}\)) x (2008 : 2)] = ( \(\frac{2008}{1}\)+ \(\frac{1}{2008}\)) : (\(\frac{1}{2}\)+ \(\frac{1}{2009}\))
A = \(\frac{\text{2008 x2008 + 1}}{2008}\)x \(\frac{2x2009+2}{2x2009}\)
A = 2008
24 tháng 6 2015
tử là M mẫu là N ta dc
\(M=2008+\frac{2007}{2}+...+\frac{1}{2008}\)
\(=\left(1+...+1\right)+\frac{2007}{2}+...+\frac{1}{2008}\)
\(=\frac{2009}{2}+...+\frac{2009}{2008}+\frac{2009}{2009}\)
\(=2009\left(\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}\right)\)
vậy ta có
\(A=\frac{M}{N}=\frac{2009\left(\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}}\)\(=2009\)
NT
29 tháng 3 2015
\(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...+\left(1+\frac{1}{2007}\right)+\left(1+\frac{1}{2008}\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)
Đặt S=\(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2008}}\)
5S=\(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2007}}\)
5S-S=\(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2007}}\)-\(\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2008}}\right)\)
4S=\(1-\frac{1}{5^{2008}}\)
=> S=\(\frac{1-\frac{1}{5^{2008}}}{4}\)