CMR :
\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
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NT
27 tháng 8 2016
\(E=1-\frac{1}{2^2}-\frac{1}{3^2}-..........-\frac{1}{2004^2}\)
\(E=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+..........+\frac{1}{2014^2}\right)\)
Ta có : \(E< 1-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{2003.2004}\right)\\ \)
Đặt A= \(1-\left(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2003.2004}\right)\\ =>A=1-\left(1-\frac{1}{2004}\right)\\ =>A=1-\frac{2003}{2004}\\ =>A=\frac{1}{2004}\)
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NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2020
Xét tổng:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2004^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}=1-\frac{1}{2004}\)
\(\Rightarrow1-A>1-\left(1-\frac{1}{2004}\right)=\frac{1}{2004}\) (đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 2 2020
\(2^2=2.2>1.2\Rightarrow\frac{1}{2^2}< \frac{1}{1.2}\)
Tương tự \(\frac{1}{3^2}< \frac{1}{2.3}\) ........
Chứng tỏ rằng :\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
8
21 tháng 11 2018
Đề: cmr: B = 1 - 1/22 - 1/32 - 1/42 -...-1/20042 > 1/2004 ( bn có ghi nhầm đề ko z)
Bài làm
ta có: \(\frac{1}{2^2}>\frac{1}{1.2};\frac{1}{3^2}>\frac{1}{2.3};\frac{1}{4^2}>\frac{1}{3.4};...;\frac{1}{2004^2}>\frac{1}{2003.2004}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)= 2003/2004
\(\Rightarrow B=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}\right)>1-\frac{2003}{2004}=\frac{1}{2004}\)
=> đpcm
T
16 tháng 2 2019
@I don't need you: Hey \(\frac{1}{2^2}>\frac{1}{1.2}\Leftrightarrow0.25>0.5?!?\)
\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)
Giải
Có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2004^2}< \frac{1}{2003.2004}\)
\(B=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}\right)\)
\(>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\right)\)
\(=1-\left(1-\frac{1}{2004}\right)=\frac{1}{2004}\) (đpcm)
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)
\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}\)
\(=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}\right)>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\right)\)
\(>1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\right)\)
\(>1-\left(1-\frac{1}{2004}\right)\)
\(>1-1+\frac{1}{2004}\)
\(>\frac{1}{2004}\left(đpcm\right)\)