CM
\(\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)
K
Khách
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28 tháng 1 2018
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+........+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
...................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}=\dfrac{6}{25}\)
Mà \(\dfrac{1}{6}< \dfrac{5}{26}< \dfrac{1}{4}\)
Mà \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+.........+\dfrac{1}{100^2}< \dfrac{6}{25}\)
\(\Leftrightarrow\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{100^2}< \dfrac{1}{4}\left(đpcm\right)\) \(\left(1\right)\)
29 tháng 4 2017
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
30 tháng 4 2017
A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
CB
1
MV
19 tháng 1 2018
\(S=\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\\ 25S=25\left(\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\right)\\ 25S=1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2016}}\\ 25S-S=\left(1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2016}}\right)-\left(\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\right)\\ 24S=1-\dfrac{1}{5^{2018}}< 1\\ \Rightarrow S< \dfrac{1}{24}\)
21 tháng 5 2022
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)
=>-10<x<-13/7
hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)
\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)
mà x là số nguyên
nên \(x\in\left\{-1;0\right\}\)
N
3 tháng 3 2019
*Có: \(A=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)\(>\dfrac{1}{4}-\dfrac{1}{101}=\dfrac{97}{404}\)\(=\dfrac{970}{4040}\)
Có: \(\dfrac{1}{5}=\dfrac{808}{4040}\)
\(\Rightarrow\dfrac{1}{5}< A\)
*Có: \(A=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)\(=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{5}< A< \dfrac{1}{3}\)
Ta có :
+) \(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
+) \(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
.......................
+) \(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+............+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+......+\dfrac{1}{100}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\) \(\left(1\right)\)
Lại có :
+) \(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)
+) \(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)
.....................
+) \(\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.........+\dfrac{1}{100.101}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{100}-\dfrac{1}{101}\)
\(\Leftrightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+........+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}>\dfrac{1}{5}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\)
Lại có \(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)