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
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{4019}{2009^2.2010^2}\)
chứng minh A < 1
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{2010^2-2009^2}{2009^2.2010^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}=1-\frac{1}{2010^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{2010^2-2009^2}{2009^2.2010^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}=1-\frac{1}{2010^2}\)
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
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+.....+\frac{4019}{2009^2.2010^2}\)
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
Chúc bạn học tốt!
CMR :
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+......+\dfrac{4019}{2009^2.2010^2}< 1\)
Ta có:
(n+1)2-n2=2n+1=n+(n+1)
=> A=\(\frac{2+1}{2^21^2}+\frac{2+3}{2^23^2}+... +\frac{2009+2010}{2009^22010^2}=1-\frac{1}{2^2}+\frac{1}{2^2} -\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2} <1 \)
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
chứng minh rằng 3/1^2.2+5/2^2.3^2+7/3^2.4^2+...+2013/1006^2.1007^2<1
Chứng minh rằng :
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\)
Chứng minh rằng: 3/1^2.2^2 + 5/2^2.3^2 + 7/3^2.4^2 + ... + 4031/2015^2.2016^2 < 1
Ta có: \(\frac{3}{1^2.2^2}=\frac{1}{1^2}-\frac{1}{2^2}\); \(\frac{5}{2^2.3^2}=\frac{1}{2^2}-\frac{1}{3^2}\); \(\frac{7}{3^2.4^2}=\frac{1}{3^2}-\frac{1}{4^2}\);....; \(\frac{4031}{2015^2.2016^2}=\frac{1}{2015^2}-\frac{1}{2016^2}\)
=> \(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2015^2}-\frac{1}{2016^2}\)
=> \(A=1-\frac{1}{2016^2}< 1\)
=> A < 1