CMR :
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+......+\dfrac{4019}{2009^2.2010^2}< 1\)
Những câu hỏi liên quan
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+.....+\frac{4019}{2009^2.2010^2}\)
3/1^2.2^2+5/2^2.3^2+7/3^2.4^2+...+4019/2009^2.2010^2
=3/1.4+5/4.9+7/9.16+...+4019/4036081.4040100
= 1/1-1/4+1/4-1/9+1/9-1/16+...+1/4036081-1/4040100
= 1/1-1/4040100
= 1-1/4040100 < 1
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a)Cho A dfrac{2015}{2016}+dfrac{2016}{2017}+dfrac{2017}{2018}+dfrac{2018}{2019}+dfrac{2019}{2020}+dfrac{2021}{2015}
Chứng minh A6
b)Cho Cdfrac{1}{3}+dfrac{1}{3^2}+dfrac{1}{3^3}+....+dfrac{1}{3^{2010}}
Chứng minh rằng C1
Cho Ddfrac{1}{1^2.2^3}+dfrac{5}{2^2.3^3}+dfrac{7}{3^2.4^2}+.....+dfrac{4019}{2009^2.2010^2}
Chứng minh rằng D1
mấy bạn giúp mình nha. Mình cần gấp lắm TT^TT
Đọc tiếp
a)Cho A= \(\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2021}{2015}\)
Chứng minh A>6
b)Cho C=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+....+\dfrac{1}{3^{2010}}\)
Chứng minh rằng C<1
Cho D=\(\dfrac{1}{1^2.2^3}+\dfrac{5}{2^2.3^3}+\dfrac{7}{3^2.4^2}+.....+\dfrac{4019}{2009^2.2010^2}\)
Chứng minh rằng D<1
mấy bạn giúp mình nha. Mình cần gấp lắm TT^TT
mấy bạn ơi câu b) là chứng minh C<\(\dfrac{1}{2}\)nha
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\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{4019}{2009^2.2010^2}\)
chứng minh A < 1
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{2010^2-2009^2}{2009^2.2010^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}=1-\frac{1}{2010^2}\)
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\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{2010^2-2009^2}{2009^2.2010^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}=1-\frac{1}{2010^2}\)
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chứng minh
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+..........+\frac{4019}{2009^2.2010^2}\)
<1
Chứng minh rằng:
a)3/1^2.2^2 + 5/2^2.3^2 + 7/3^2.4^2 + ... + 4019/2009^2.2010^2 < 1
b) (1+ 1/3 ).(1+ 1/8).(1+ 1/15). ... .(1+ 1/n^2+ 2n) < 2
chứng tỏ rằng:
a)3/1^2.2^2 + 5/2^2.3^2 + 7/3^2.4^2 + ... + 4019/ 2009^2.2010^2 < 1
b) (1+ 1/3 ).(1+ 1/8).(1+ 1/15). ... .(1+ 1/n^2+ 2n) < 2
Cmr: \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
Ta có:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
= \(\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
= \(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
= \(1-\dfrac{1}{10^2}\)
Mà \(1-\dfrac{1}{10^2}< 1\) nên:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) < 1 (đpcm).
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cmr \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{4027}{2013^2.2014^2}< 1\)
cmr
\(\dfrac{1}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{4017}{2013^2.2014^2}< 1\)
Ta thấy 32-22=5; 42-32=7;......;20142-20132=(2014-2013)(2014+2013)=4017
=> VT=1/4+1/4-1/9+1/9-1/16+1/16-......-1/20132+1/20132-1/20142
=1/4+1/4-1/2014=1/2-1/20142<1/2<1
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