\(CM:A=\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{2007^2}< \frac{1003}{2008}\)
Những câu hỏi liên quan
Cho:A=\(\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{2007^2}\) Cm:A<\(\frac{1003}{2008}\)
Chứng minh \(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2007^2}
Ta thấy: 32>32-1=(3-1).(3+1)=2.4
52>52-1=(5-1).(5+1)=4.6
72>72-1=(7-1).(7+1)=6.8
…………………………
20072>20072-1=(2007-1).(2007+1)=2006.2008
=> \(\frac{2}{3^2}
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\(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+.....+\frac{2}{2007^2}\)
CMR A<1003/2008
Cho A = \(\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2007^2}\)Chứng minh rằng : A< 1003/2008
\(A=\frac{2}{3^2}+\frac{2}{5^2}+.......+\frac{2}{2007^2}\)
\(A=2.\left(\frac{1}{3.3}+\frac{1}{5.5}+......+\frac{1}{2007.2007}\right)\)
\(A< 2.\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2006.2007}\right)\)
\(A< 2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2006}-\frac{1}{2007}\right)\)
\(A< 2.\left(\frac{1}{2}-\frac{1}{2007}\right)\)
\(A< 2.\frac{2005}{4014}\)
\(A< \frac{2005}{2007}\)
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Ta thấy
2/(3x3) < 2/(2x4) = 1/2 – 1/4
2/(5x5) < 2/(4x6) = 1/4 – 1/6
2/(7x7) < 2/(6x8) = 1/6 – 1/8
………
2/(2007x2007) < 2/(2006x2008) = 1/2006 – 1/12008
Nên:
A = 2/3^2 +2/5^2+2/7^2 +.....+2/2007^2 < 2/(2x4) + 2/(4x6) + …. + 2/(2006x2008) =
1/2 – 1/4 + 1/4 – 1/6 + 1/6 – 1/8 + … + 1/2006 – 1/2008 =
1/2 – 1/2008 = 1003/2008
Vậy: .....
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Cho A = \(\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+....+\frac{2}{2007^2}\)Chứng minh : A < \(\frac{1003}{2008}\)
\(A< \frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2007.2009}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}=\frac{1}{3}-\frac{1}{2009}=\frac{2006}{6027}< \frac{2006}{4016}=\frac{1003}{2008}\)Vây:.......
Cho \(B=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+.....+\frac{2}{2007^2}\). Chứng minh: A<\(\frac{1003}{2008}\)
Chứng minh rằng :
\(\frac{2}{3^2}\)+\(\frac{2}{5^2}\)+...+\(\frac{2}{2007^2}\)< \(\frac{1003}{2008}\)
Ta thấy: \(\frac{2}{3^2}=\frac{2}{3.3}< \frac{2}{2.4}=\frac{1}{2}-\frac{1}{4}\)
\(\frac{2}{5.5}< \frac{2}{4.6}=\frac{1}{4}-\frac{1}{6}\)\(;...;\frac{2}{2007.2007}< \frac{2}{2006.2008}=\frac{1}{2006}-\frac{1}{2008}\)
\(\Rightarrow\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{2007^2}< \frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2006}-\frac{1}{2008}\)
Ta có:\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2006}-\frac{1}{2008}=\frac{1}{2}-\frac{1}{2008}=\frac{1004-1}{2008}=\frac{1003}{2008}\)
\(\Rightarrow\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{2007^2}< \frac{1003}{2008}\)(đpcm)
K mình nè!
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CHO A =\(\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2007^2}\)
CHUNG MINH A <\(\frac{1003}{2008}\)
CM
\(A=\frac{2}{3^2}+\frac{2}{5^2}+...+\frac{2}{2007^2}\) <\(\frac{1003}{2008}\)
\(B=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2006^2}<\frac{334}{2007}\)
\(S=\frac{1}{5^2}+\frac{1}{9^2}+...+\frac{1}{409^2}<\frac{1}{12}\)
Đặt B=\(\frac{2}{4^2}+\frac{2}{6^2}+\frac{2}{8^2}+....+\frac{2}{2008^2}\)
=> A+B= 2\(\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2007^2}+\frac{1}{2008^2}\right)\) <2 \(\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{2006\cdot2007}+\frac{1}{2007\cdot2008}\right)\)
=2\(\left(\frac{1}{2}-\frac{1}{2008}\right)\)=\(\frac{2006}{2008}\)
mà A<B=>A+A<A+B=2006/2008
=>A<1003/2008
mấy câu kia cũng tương tự, mình làm biếng quá
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