Cho B = \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{3^3}\)+ \(\frac{1}{3^4}\)+...+ \(\frac{1}{3^{2004}}\)+ \(\frac{1}{3^{2005}}\) . CMR : B < \(\frac{1}{2}\)
K
Khách
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Những câu hỏi liên quan
11 tháng 7 2018
a) \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2015}}\)
\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)
20 tháng 7 2016
Mẫu số = 2004/1 + 2003/2 + 2002/3 + ... + 1/2004
= (1 + 1 + ... + 1) + 2003/2 + 2002/3 + ... + 1/2004
2004 số 1
= (1 + 2003/2) + (1 + 2002/3) + ... + (1 + 1/2004) + 1
= 2005/2 + 2005/3 + ... + 2005/2004 + 2005/2005
= 2005 × (1/2 + 1/3 + ... + 1/2004 + 1/2005)
=> B = 1/2005
S
20 tháng 7 2016
Mẫu số = 2004/1 + 2003/2 + 2002/3 + ... + 1/2004
= (1 + 1 + ... + 1) + 2003/2 + 2002/3 + ... + 1/2004
2004 số 1
= (1 + 2003/2) + (1 + 2002/3) + ... + (1 + 1/2004) + 1
= 2005/2 + 2005/3 + ... + 2005/2004 + 2005/2005
= 2005 × (1/2 + 1/3 + ... + 1/2004 + 1/2005)
=> B = 1/2005
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2019
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< \sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(P=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2005\sqrt{2004}}\)
\(\Rightarrow P< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(\Rightarrow P< 2\left(1-\frac{1}{\sqrt{2005}}\right)< 2.1=2\)
LD
CMR \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}< 2\)
1
AH
Akai Haruma
Giáo viên
3 tháng 5 2019
Lời giải:
Xét số hạng tổng quát \(\frac{1}{(n+1)\sqrt{n}}\):
\(\frac{1}{(n+1)\sqrt{n}}=\frac{(n+1)-n}{(n+1)\sqrt{n}}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}.\sqrt{n(n+1)}}\)
\(< \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\frac{\sqrt{n+1}+\sqrt{n}}{2}.\sqrt{n(n+1)}}\)
\(\Leftrightarrow \frac{1}{(n+1)\sqrt{n}}< 2.\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Cho $n=1,2,....,2004$
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{2005\sqrt{2004}}< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{2005\sqrt{2004}}< 2(1-\frac{1}{\sqrt{2005}})< 2\) (đpcm)
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
13 tháng 1 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 + 1/3^3 + ..... + 1/3^2005 )
= 1 - 1/3^2005 < 1
=> B < 1 : 2 = 1/2
=> ĐPCM
Tk mk nha
13 tháng 1 2018
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}< 1\)
\(\Rightarrow B< \frac{1}{2}\)
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 = \(\dfrac{1}{2}\)-\(\dfrac{1}{\dfrac{3^{2005}}{2}}\) < \(\dfrac{1}{2}\)