Bài 1 : Cho N =\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)
Hãy chứng minh rằng N<1
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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
30 tháng 4 2018
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
16 tháng 3 2019
\(N=\) \(\frac{1}{2^2}\) \(+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2008.2009}+\frac{1}{2009.2010}\)
\(N< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2010}\)
\(N< 1-\frac{1}{2010}\)
\(N< \frac{2009}{2010}< 1\)
\(\Rightarrow N< 1\)
23 tháng 10 2016
a) \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
\(=\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^2\left(n^2+2n+1+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=>đpcm
b) Từ công thức trên ta có:
\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=> \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(=2010+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(2010+\left(1-\frac{1}{2011}\right)=2010+\frac{2010}{2011}=2010\frac{2010}{2011}\)
NC
29 tháng 8 2020
a) Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; \(\frac{1}{4^2}< \frac{1}{3.4}\) ; ... ; \(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(Vt< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
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
câu này hay thế!