1. Tính:
\(S=\frac{1+2^2+2^3+...+2^{2008}}{1-2^{2009}}\); \(P=\frac{1+4+7+11+...+127}{\frac{2}{4}.\frac{6}{9}.\frac{12}{16}...\frac{420}{400}}\)
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\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
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:\(S=\frac{2^{2009}-1}{1-2^{2009}}=-1\)
Tính tổng S=\(\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
Làm giúp mk bài này nha!Cảm ơn mn nhiều:3
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
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
mình ko chắc đâu đó nha,bài này mình chỉ làm có mấy lần à,sai thì cho mình xin lỗi nhé T_T
Tử = 1+2+2^2+2^3+...+2^2008
2Tử = 2+2^2+2^3+...+2^2009
=> 2Tử-Tử=2^2009-1
S= (2^2009-1)/(1-2^2009)=-1
Cho A = 1 + 2 + 22 + 23 + ... + 22008
-> 2A = 2 + 22 + 23 + 24 +...+ 22009
-> 2A - A = ( 2 + 22 + 23 + 24 +...+ 22009 ) - ( 1 + 2 + 22 + 23 + ... + 22008 )
-> A = \(2^{2009}-1=-\left(1-2^{2009}\right)\)
S = \(\frac{-\left(1-2^{2009}\right)}{1-2^{2009}}\)=-1