N=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)
tính N<1
K
Khách
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Những câu hỏi liên quan
HT
28 tháng 2 2018
a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1
= 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011
= 2011(1/2+1/3+1/4+...+1/2011)
Ta có: B= 1/2+1/3+1/4+...+1/2011
suy ra A/B= 2011
13 tháng 9 2016
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+...+\left(\frac{1}{2010}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}\right)}\)
\(A=\frac{1}{2011}\)
LA
7 tháng 2 2018
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Ta có: \(C=\frac{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}\)
Đặt \(A=\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}\)
\(A=\frac{2010}{1}+1+\frac{2009}{1}+1+\frac{2008}{1}+1+...+\frac{1}{2010}+1-2010\)
\(=\frac{2011}{1}+\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}-\frac{2011.2010}{2011}\)
\(=2011\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}-1\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\)
Ta có: \(C=\frac{A}{B}=2011\)(lấy A-B)
6 tháng 2 2018
Ta có :
\(2010A=\dfrac{2010^{2012}+2010}{2010^{2012}+1}=\dfrac{2010^{2012}+1+2009}{2010^{2012}+1}=1+\dfrac{2009}{2010^{2012}+1}\)
\(2010B=\dfrac{2010^{2011}+2010}{2010^{2011}+1}=\dfrac{2010^{2011}+1+2009}{2010^{2011}+1}=1+\dfrac{2009}{2010^{2011}+1}\)
Vì \(1+\dfrac{2009}{2010^{2012}+1}< 1+\dfrac{2009}{2010^{2011}+1}\Rightarrow A< B\)
~ Học tốt ~
27 tháng 6 2019
Xét N :
N = \(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+...+\(\frac{1}{2009.2009}\)+\(\frac{1}{2010.2010}\)
Ta có :
\(\frac{1}{2.2}\)< \(\frac{1}{1.2}\)
\(\frac{1}{3.3}\)< \(\frac{1}{2.3}\)
...
\(\frac{1}{2009.2009}\)<\(\frac{1}{2008.2009}\)
\(\frac{1}{2010.2010}\)<\(\frac{1}{2019.2010}\)
Cộng vế theo vế của các bất đẳng thức trên , ta có :
\(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+...+\(\frac{1}{2009.2009}\)+\(\frac{1}{2010.2010}\) < \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2008.2009}\)+\(\frac{1}{2019.2010}\)
=> N < 1 - \(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2009}\)-\(\frac{1}{2010}\)
=> N < 1 - \(\frac{1}{2010}\)<1
=> N < 1
DD
Đoàn Đức Hà
Giáo viên
17 tháng 6 2021
Tổng quát: \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=1+\frac{\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}=1+\frac{1}{n^2\left(n+1\right)^2}+\frac{2}{n\left(n+1\right)}\)
\(=\left(1+\frac{1}{n\left(n+1\right)}\right)^2=\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2\)
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\left|1+\frac{1}{n}-\frac{1}{n+1}\right|=1+\frac{1}{n}-\frac{1}{n+1}\)
Áp dụng ta được:
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2009^2}+\frac{1}{2010^2}}\)
\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2009}-\frac{1}{2010}\)
\(=2008+\frac{1}{2}-\frac{1}{2010}\)
\(=2008\frac{502}{1005}\)
15 tháng 4 2017
Ta có: A=\(\frac{1}{2011}+\frac{2}{2010}+\frac{3}{2009}+...+\frac{2009}{3}+\frac{2010}{2}+\frac{2011}{1}\)
=> A=\(\frac{2012-2011}{2011}+\frac{2012-2010}{2010}+...+\frac{2012-2}{2}+\frac{2012-1}{1}\)
=>A=\(\frac{2012}{2011}-1+\frac{2012}{2010}-1+...+\frac{2012}{2}-1+2012-1\)
=>A=\(2012\cdot\left(\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{2}\right)+1\)
=> A= \(2012\cdot\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{2}\right)\)
ko biết có đúng hay ko nựa sai thì bỏ qua nha ^^
Ta có
1/2^2<1/1.2
1/3^2<1/2.3
......
1/2009^2<1/2008.2009
1/2010^2<1/2009.2010
=>1/2^2+1/3^2+...+1/2010^2<1/1.2+1/2.3+....+1/2009.2010
=>N<1/1.2+1/2.3+....+1/2009.2010
=>N<1-1/2010
=>N<2009/2010<1
Vậy N<1
\(N=\) \(\frac{1}{2^2}\) \(+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2008.2009}+\frac{1}{2009.2010}\)
\(N< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2010}\)
\(N< 1-\frac{1}{2010}\)
\(N< \frac{2009}{2010}< 1\)
\(\Rightarrow N< 1\)