Cho \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2005}}\)
CMR:\(B< \frac{1}{2}\)
K
Khách
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DT
3
LK
7 tháng 3 2018
ta có \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}.\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)
\(\Leftrightarrow3B-B=1+\frac{1}{3}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^2}+...+\frac{1}{3^{2004}}-\frac{1}{3^{2004}}-\frac{1}{3^{2005}}\)
\(\Leftrightarrow2B=1-\frac{1}{3^{2005}}\) \(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\left(đpcm\right)\)
7 tháng 3 2018
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
MK
3 tháng 2 2017
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)
5 tháng 3 2018
bạn ơi mình sửa cho bạn nè!
B=(1-\(\dfrac{1}{3^{2005}}\)) :2 = \(\dfrac{1}{2}\)-\(\dfrac{1}{\dfrac{3^{2005}}{2}}\) < \(\dfrac{1}{2}\)
HQ
1
T
11 tháng 10 2018
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:
\(3B-B=2B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)
\(2B=1-\frac{1}{3^{2005}}< 1\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}^{\left(đpcm\right)}\)
3 tháng 1 2017
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}
\)
\(2B=1-\frac{1}{3^{2004}}\)
\(B=\frac{1}{2}-\frac{1}{2\cdot3^{2004}}\)
Do đó B<\(\frac{1}{2}\)
chúc thành công
10 tháng 2 2016
1.
A=19^5^1^8^9^0+2^9^1^9^6^9
Ta luôn có 1a=1 với a là số nguyên dương
=>19^5^1^8^9^0=195 và 2^9^1^9^6^9=29
=>A=195+29=(192)2.19+(24)2.2=(...1)2.19+(...6)2.2=...1.19+...6.2=...1
Vậy A có tận cung là 1.
2.
B=1/3+1/32+...+1/32005
3B=1+1/3+1/32+...+1/32004
3B-B=1-1/32005
2B=1-1/32005<1
=>2B<1=>B<1/2
Vậy B<1/2.
.
.
10 tháng 2 2016
1) Ta có:
\(19^{5^{1^{8^{9^0}}}}+2^{9^{1^{9^{6^9}}}}=19^{5^1}+2^{9^1}\)
Mà 195=194+1=...1.19=...19
29=22.4+1=...6 .2=...2
=>A=...19 + ...2= ...1
Vậy A có chữ số tận cùng là 1
3 tháng 3 2016
\(\Rightarrow3B=3+\frac{1}{3^1}+\frac{1}{3^2}+....+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=3-\frac{1}{3^{2005}}\Rightarrow B=\left(3-\frac{1}{3^{2005}}\right):2\)
\(\Rightarrow\left(3-\frac{1}{3^{2005}}\right):2<\frac{1}{2}\Rightarrow B<\frac{1}{2}\)
LC
3 tháng 3 2016
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
11 tháng 7 2018
a) \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2015}}\)
\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2019
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< \sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(P=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2005\sqrt{2004}}\)
\(\Rightarrow P< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(\Rightarrow P< 2\left(1-\frac{1}{\sqrt{2005}}\right)< 2.1=2\)
Ta có \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\Rightarrow\frac{1}{3}.B=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)
\(\Rightarrow B-\frac{1}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)
\(\frac{2}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)
\(B=\left(\frac{1}{3}-\frac{1}{3^{2006}}\right):\frac{2}{3}\)
\(B=\frac{1}{3}:\frac{2}{3}-\frac{1}{3^{2006}}:\frac{2}{3}=\frac{1}{2}-\frac{1}{2.3^{2005}}< \frac{1}{2}\)