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Có B=\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+...+\(\frac{1}{3^{2004}}\)+\(\frac{1}{3^{2005}}\)
=>3B=3.(\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\))
=>3B=1+\(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
=>3B-B=(1+\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\))-(\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\))
=>2B=\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-....-\frac{1}{3^{2004}}-\frac{1}{3^{2005}}\)
=>2B=1-\(\frac{1}{3^{2005}}\)
=>B=(\(1-\frac{1}{3^{2005}}\)):2
Mà \(\left(1-\frac{1}{3^{2005}}\right)< \frac{1}{2}\)=>\(\left(1-\frac{1}{3^{2005}}\right):2< \frac{1}{2}\)
=>B<\(\frac{1}{2}\)(đpcm)
bạn ơi mình sửa cho bạn nè!
B=(1-\(\dfrac{1}{3^{2005}}\)) :2 = \(\dfrac{1}{2}\)-\(\dfrac{1}{\dfrac{3^{2005}}{2}}\) < \(\dfrac{1}{2}\)
Có :
3B = 1 + 1/3 + 1/3^2 + .... + 1/3^2004
2B = 3B - B = ( 1 + 1/3 + 1/3^2 + ..... + 1/3^2004 ) - ( 1/3 + 1/3^2 + 1/3^3 + ..... + 1/3^2005 )
= 1 - 1/3^2005 < 1
=> B < 1 : 2 = 1/2
=> ĐPCM
Tk mk nha
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}< 1\)
\(\Rightarrow B< \frac{1}{2}\)
\(\Rightarrow3B=3+\frac{1}{3^1}+\frac{1}{3^2}+....+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=3-\frac{1}{3^{2005}}\Rightarrow B=\left(3-\frac{1}{3^{2005}}\right):2\)
\(\Rightarrow\left(3-\frac{1}{3^{2005}}\right):2<\frac{1}{2}\Rightarrow B<\frac{1}{2}\)
3B=1+1/3+1/32+...+1/32004
3B-B=1-1/32005
2B=1-1/32005
B=1/2-1/(32005.2)
Vậy B <1/2
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(P=\frac{1}{5}-\frac{2}{3}=\frac{3-10}{15}=\frac{-7}{15}\)
khó thế mk chịu r
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