A=1/2^2+1/3^2+1/4^2+...+1/9^2 chứng minh 2/5 < A <8/9
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NM
1
24 tháng 6 2015
ta có A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\) < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
= \(1-\frac{1}{9}\)
= \(\frac{8}{9}\)
suy ra A < \(\frac{8}{9}\)
ta có A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)j> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
= \(\frac{1}{2}-\frac{1}{10}\)
= \(\frac{2}{5}\)
suy ra A >\(\frac{2}{5}\)
NT
3
AK
15 tháng 4 2018
Ta có :
\(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};...;\frac{1}{9^2}>\frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{5}{10}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{4}{10}=\frac{2}{5}\left(1\right)\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{9^2}< \frac{1}{8.9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< 1-\frac{1}{9}=\frac{8}{9}\left(2\right)\)
Từ ( 1 ) , ( 2 ) => ĐPCM
Chúc bạn học tốt !!!
AK
15 tháng 4 2018
Đề sai bạn nhé :
Đề đúng :
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
CM : \(\frac{2}{5}< A< \frac{8}{9}\)
KH
0
TL
2
N
24 tháng 3 2017
Ta có:\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
Lại có \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
Mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{2}{5}\)
Vậy \(\frac{2}{5}< S< \frac{8}{9}\)
BT
24 tháng 3 2017
S< 1/1.2+1/2.3+1/3.4+...+1/8.9 = 1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9=1-1/9=8/9
=> S < 8/9
S> 1/2.3+1/3.4+1/4.5+...+1/9.10=1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10=1/2-1/10=4/10=2/5
=> S > 2/5
Đs: 2/5 < S < 8/9
NB
5
JK
26 tháng 7 2019
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}< \frac{1}{2\cdot3}\); \(\frac{1}{4^2}< \frac{1}{3\cdot4}\); ....; \(\frac{1}{9^2}< \frac{1}{8\cdot9}\)
\(\Rightarrow S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow S< 1-\frac{1}{9}\)
\(\Rightarrow S< \frac{8}{9}\) (1)
\(\frac{1}{2^2}>\frac{1}{2\cdot3};\frac{1}{3^2}>\frac{1}{3\cdot4};\frac{1}{4^2}>\frac{1}{4\cdot5};...;\frac{1}{9^2}>\frac{1}{9\cdot10}\)
\(\Rightarrow S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow S>\frac{2}{5}\) (2)
(1)(2) => 2/5 < S < 8/9
26 tháng 7 2019
\(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}< \frac{1}{a^2}\)
\(\frac{1}{a}-1-\frac{1}{a}=-1< \frac{1}{a^2}\) Vì \(\frac{1}{a^2}>0;-1< 0\)
Khi đó thì ĐỀ SAI
AD
1
HT
9 tháng 8 2015
A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\)
A < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
A < \(1-\frac{1}{9}\)
A < \(\frac{8}{9}\)
A > \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
A > \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
A > \(\frac{1}{2}-\frac{1}{10}\)
A > \(\frac{2}{5}\)
KL: \(\frac{2}{5}\)< A < \(\frac{8}{9}\) (đpcm)
TH
1
15 tháng 8 2018
\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2};...;\frac{1}{9\cdot9}< \frac{1}{8\cdot9}\)
\(\Rightarrow S=\frac{1}{2^2}+...+\frac{1}{9^2}< \frac{1}{1\cdot2}+...+\frac{1}{8\cdot9}=1-\frac{1}{2}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)
\(\frac{1}{2\cdot2}>\frac{1}{2\cdot3};...;\frac{1}{9\cdot9}>\frac{1}{9\cdot10}\)
\(\Rightarrow S=\frac{1}{2^2}+...+\frac{1}{9^2}>\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}=\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(2\right)\)
Từ (1)(2) => đpcm
A = 1 / 2.2 + 1 / 3.3 + 1 / 4.4 + .... + 1 / 9.9
A < 1/1.2 + 1/2.3 + .....+ 1/8.9
A < 1 - 1/2 + 1/2 - 1/3 + ......+ 1/8 - 1/9
A < 1 - 1/9
=> A < 8/9 (1)
Mặt khác ta có:
A > 1/2.3 + 1/3.4 +.....+ 1/9.10
A > 1/2 - 1/3 + 1/3 - 1/4 +.......+ 1/9 - 1/10
A > 1/2 - 1/10
A > 4/10
=> A > 2/5 (2)
Từ (1) và (2) => 8/9 > A > 2/5
\(\frac12-\frac13=\frac{1}{2\cdot3}<\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac13-\frac14=\frac{1}{3\cdot4}<\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac19-\frac{1}{10}=\frac{1}{9\cdot10}<\frac{1}{9^2}<\frac{1}{8\cdot9}=\frac18-\frac19\)
Do đó: \(\frac12-\frac13+\frac13-\frac14+\cdots+\frac19-\frac{1}{10}<\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{9^2}<1-\frac12+\frac12-\frac13+\cdots+\frac18-\frac19\)
=>\(\frac12-\frac{1}{10}
=>\(\frac25