B = 1/3 + 1/3^2 + 1/3^3 +....+ 1/3^2005
Cm B<1/2
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DM
0
PM
1
LM
2
XO
4 tháng 9 2020
Ta có : B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\)
=> 3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)
Khi đó 3B - B = \(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
=> 2B = \(1-\frac{1}{3^{2005}}\)
=> B = \(\frac{1}{2}-\frac{1}{3^{2005}.2}< \frac{1}{2}\left(\text{ĐPCM}\right)\)
NN
4 tháng 9 2020
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+........+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)
Vì \(1-\frac{1}{3^{2005}}< 1\)\(\Rightarrow\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)
hay \(B< \frac{1}{2}\)( đpcm )
VT
1
14 tháng 1 2016
Ta có:3B\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+...+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}\)
B=\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+..+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}+\frac{1}{3}^{2005}\)
\(\Rightarrow\)2B=1-\(\frac{1}{3}^{2005}\)
\(\Rightarrow\)B=\(\frac{1-\frac{1}{3}^{2005}}{2}\)
\(\Rightarrow\)B=\(\frac{1-\frac{1}{3}^{2005}}{2}<\frac{1}{2}\)
\(\Rightarrow\)B<\(\frac{1}{2}\)
LH
3
6 tháng 3 2017
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Leftrightarrow2B=3\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(\Leftrightarrow2B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Leftrightarrow2B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{3^{2005}}\)
\(\Leftrightarrow B=1-\frac{1}{3^{2005}}< \frac{1}{2}\)
Vậy \(B< \frac{1}{2}\) (Đpcm)
6 tháng 3 2017
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\ \)
\(C=3B=1+\dfrac{1}{3}+..+\dfrac{1}{3^{2004}}\)
\(C-B=1-\dfrac{1}{3^{3005}}\)
\(B=\dfrac{1}{2}-\dfrac{1}{2.3^{2005}}< \dfrac{1}{2}\)
TP
1
HT
1 tháng 6 2016
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{2005^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2004.2005}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2004}-\frac{1}{2005}\)
\(=1-\frac{1}{2005}=\frac{2004}{2005}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{2005^2}< \frac{2004}{2005}\left(\text{đ}pcm\right)\)
VH
1
27 tháng 4 2017
Ta có :
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(2B=1-\frac{1}{3^{2005}}< 1\)
\(\Rightarrow\frac{2B}{2}=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)
\(\Rightarrow B< \frac{1}{2}\)
NT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
21 tháng 9 2015
\(\frac{B}{3}=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)
\(\frac{2B}{3}=B-\frac{B}{3}=\frac{1}{3}-\frac{1}{3^{2006}}\)
\(2B=1-\frac{1}{3^{2005}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2005}}<\frac{1}{2}\)
QQ
5
1 tháng 8 2015
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}+\frac{1}{3^{2015}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2014}}+\frac{1}{3^{2015}}\right)\)
\(2B=1-\frac{1}{3^{2015}}\)
\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)
Mà \(1-\frac{1}{3^{2015}}<1\)
\(\Rightarrow B<\frac{1}{2}\)
Vậy ____________
DV
3 tháng 2 2017
Câu của đặng phương thảo sai rồi ở 3b-b thì là 3^2005 chứ không phải là 3^ 2015
\(\frac{B}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2006}}\)
\(\frac{2B}{3}=B-\frac{B}{3}=\frac{1}{3}-\frac{1}{3^{2006}}\Rightarrow2B=1-\frac{1}{2^{2005}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2005}}< \frac{1}{2}\)