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Ta có \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\Rightarrow\frac{1}{3}.B=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)
\(\Rightarrow B-\frac{1}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)
\(\frac{2}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)
\(B=\left(\frac{1}{3}-\frac{1}{3^{2006}}\right):\frac{2}{3}\)
\(B=\frac{1}{3}:\frac{2}{3}-\frac{1}{3^{2006}}:\frac{2}{3}=\frac{1}{2}-\frac{1}{2.3^{2005}}< \frac{1}{2}\)
ta có \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}.\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)
\(\Leftrightarrow3B-B=1+\frac{1}{3}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^2}+...+\frac{1}{3^{2004}}-\frac{1}{3^{2004}}-\frac{1}{3^{2005}}\)
\(\Leftrightarrow2B=1-\frac{1}{3^{2005}}\) \(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\left(đpcm\right)\)
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 + ...... + 1/3^2004 )
= 1 - 1/3^2004 < 1
=> B < 1/2
Tk mk nha
Sửa đề: Cho \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\). CMR: \(B< \frac{1}{2}\)
Ta có: \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\). Lại có:
\(3B-B=2B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)
\(2B=1-\frac{1}{3^{2005}}< 1\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}^{\left(đpcm\right)}\)
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
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}< 1\)
\(\Rightarrow B< \frac{1}{2}\)
a, \(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}\)
\(C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)(đpcm)
b, Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{397}{3^{100}}\)
\(A=\frac{3}{4}-\frac{397}{4.3^{100}}< \frac{3}{4}\)(đpcm)
B=1/3+1/32+...+1/32017 <1/2
3B=1+1/3+1/32+...1/32016 <1/2
3B-B=(1+1/3+...+1/32016) - (1/3+1/32+...+1/32017)
2B=1-(1/32017)
2B=(32017-1) phần (32017)=>B=(32017-1):2 phần (32017)
Vậy ..........................
\(\Rightarrow3B=3+\frac{1}{3^1}+\frac{1}{3^2}+....+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=3-\frac{1}{3^{2005}}\Rightarrow B=\left(3-\frac{1}{3^{2005}}\right):2\)
\(\Rightarrow\left(3-\frac{1}{3^{2005}}\right):2<\frac{1}{2}\Rightarrow B<\frac{1}{2}\)
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
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}} \)
\(2B=1-\frac{1}{3^{2004}}\)
\(B=\frac{1}{2}-\frac{1}{2\cdot3^{2004}}\)
Do đó B<\(\frac{1}{2}\)
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