Edited by jb7185, 26-12-2012 - 22:36.
Cho $f(x)=4^x.(4^x+2)^{-1}$ tính tổng:$S=f(\frac{1}{2010})+f(\frac{2}{2010})+...+f(\frac{2009}{2010})$
Started By jb7185, 26-12-2012 - 22:34
#1
Posted 26-12-2012 - 22:34
jb7185
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Cho $f(x)=4^x.(4^x+2)^{-1}$ tính tổng:$S=f(\frac{1}{2010})+f(\frac{2}{2010})+...+f(\frac{2009}{2010})$
#2
Posted 29-12-2012 - 15:44
maitienluat
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Dề dàng kiểm tra được rằng $f(1-x)=\frac{2}{4^{x}+2}\Rightarrow f(x)+f(1-x)=1$
Từ đó $S=f(\frac{1}{2010})+f(\frac{2009}{2010})+f(\frac{2}{2010})+f(\frac{2008}{2010})+...+f(\frac{1005}{2010})=1004+f(\frac{1}{2})=\frac{2009}{2}$
Từ đó $S=f(\frac{1}{2010})+f(\frac{2009}{2010})+f(\frac{2}{2010})+f(\frac{2008}{2010})+...+f(\frac{1005}{2010})=1004+f(\frac{1}{2})=\frac{2009}{2}$
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