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B
0
Q
1
YN
8 tháng 4 2022
`Answer:`
\(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}\)
a) Ta thấy:
\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)
\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)
\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>16.\frac{1}{32}=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}\)
b) Ta thấy:
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)
\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)
\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)
\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)
\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)
MM
14
5 tháng 8 2015
Ta có S=1/2^2+1/3^2+1/4^2+...+1/9^2
<1/2²+1/2*3+1/3*4+....+1/8*9
=1/2²+1/2-1/3+1/3-1/4+....+1/8-1/9
=1/4+1/2-1/9=23/36<32/36=8/9 (♪)
Ta lại có S=1/2^2+1/3^2+1/4^2+...+1/9^2
>1/2²+1/3*4+1/4*5+....+1/9*10
=1/2²+1/3-1/4+1/4-1/5+........+1/9-1/10
=1/2²+1/3-1/10
=19/20>8/20=2/5 ( ♫)
Từ (♪)( ♫) cho ta đpcm
HT
0
MM
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
\(S=\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+...+\dfrac{1}{103^2}\)
\(\Rightarrow2S=\dfrac{2}{5^2}+\dfrac{2}{7^2}+\dfrac{2}{9^2}+...+\dfrac{2}{103^2}\)
Có:
\(\dfrac{2}{5^2}=\dfrac{2}{5.5}< \dfrac{2}{4.6}=\dfrac{1}{4}-\dfrac{1}{6}\)
\(\dfrac{2}{7^2}=\dfrac{2}{7.7}< \dfrac{2}{6.8}=\dfrac{1}{6}-\dfrac{1}{8}\)
\(\dfrac{2}{9^2}=\dfrac{2}{9.9}< \dfrac{2}{8.10}=\dfrac{1}{8}-\dfrac{1}{10}\)
...
\(\dfrac{2}{103^2}=\dfrac{2}{103.103}< \dfrac{1}{102.104}=\dfrac{1}{102}-\dfrac{1}{104}\)
\(\Rightarrow2S< \dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{102}-\dfrac{1}{104}\)
\(\Rightarrow2S< \dfrac{25}{104}\)
\(\Rightarrow S< \dfrac{25}{208}< \dfrac{5}{32}\)
\(\Rightarrow S< \dfrac{5}{32}\).
Ta có:
\(\dfrac{1}{5^2}< \dfrac{1}{4.6}\)
\(\dfrac{1}{7^2}< \dfrac{1}{6.8}\)
\(\dfrac{1}{9^2}< \dfrac{1}{8.10}\)
\(...\)
\(\dfrac{1}{103^2}< \dfrac{1}{102.104}\)
\(\Rightarrow S\)\(< \dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{102.104}\)\(\left(1\right)\)
Đặt \(A=\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{102.104}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+...+\dfrac{2}{102.104}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{102}-\dfrac{1}{104}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{104}\right)\)
\(=\dfrac{1}{2}.\dfrac{25}{104}\)
\(=\dfrac{25}{208}< \dfrac{25}{160}\)\(\left(2\right)\)
Mà \(\dfrac{25}{160}=\dfrac{5}{32}\)\(\left(3\right)\)
Từ \(\left(1\right),\left(2\right)\) và \(\left(3\right)\)
\(\Rightarrow S< \dfrac{5}{32}\)