chung minh A = 1/4 +1/16 +1/36 ... + 1/10000
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TK
0
TT
2
20 tháng 3 2016
Đặt S=1/4+1/16+1/36+...+1/10000
S= 1/4x(1+1/4+1/9+...+1/2500)
S= 1/4x(1+1/2x2+1/3x3+...+1/50x50)
S< 1/4x(1+1/1x2+1/2x3+...1/49x50)
S< 1/4x(1+1-1/2+1/2-1/3+....+1/49-1/50)
S< 1/4x(1+1-1/50)
S< 1/4x(2-1/50)<2/4(2/4=1/2)
S< 1/2
DD
20 tháng 3 2016
S=\(\frac{1}{4}\)(1+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}+...+\frac{1}{50^2}\)
S<\(\frac{1}{4}\)(1+\(\frac{1}{2.1}\)+\(\frac{1}{3.2}+...+\frac{1}{50.49}\))
S<\(\frac{1}{4}\)(1+1−\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\))
S<\(\frac{1}{4}\)(2−\(\frac{1}{50}\))<\(\frac{2}{4}\)=\(\frac{1}{2}\)(đpcm)
LG
4
KN
14 tháng 9 2020
Đề phải là \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}< \frac{1}{2}\) chứ ?
Ta có : \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)
\(=\frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{2500}\right)\)
\(=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Ta lại có : \(\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(=\frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{4}\left(2-\frac{1}{50}\right)=\frac{1}{4}.\frac{99}{50}=\frac{99}{200}\)
Mà \(\frac{99}{200}< \frac{1}{2}\)\(\Rightarrow\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}< \frac{99}{200}< \frac{1}{2}\)
\(\Rightarrow\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}< \frac{1}{2}\) ( đpcm )
NH
14 tháng 9 2020
\(\text{Đặt BT là A }\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(\text{Ta có:}\frac{1}{3^2}>\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\text{(để lại }\frac{1}{4}\text{ở đầu)}\)
\(\frac{1}{4^2}>\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)
.......
\(\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{101}\Rightarrow A=\frac{7}{12}-\frac{1}{101}=\frac{707-12}{1212}=\frac{695}{1212}>\frac{606}{1212}=\frac{1}{2}\)
\(\Rightarrow A>\frac{1}{2}\)
17 tháng 3 2017
Đặt \(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)
Ta có:
\(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)
\(\Rightarrow A=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(\Rightarrow A< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)
\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)\)
\(\Rightarrow A< \dfrac{1}{4}.\dfrac{99}{50}\)
\(\Rightarrow A< \dfrac{99}{200}< \dfrac{1}{2}\)
Vậy \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}< \dfrac{1}{2}\) (Đpcm)
17 tháng 3 2017
\(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(2-\dfrac{1}{50}\right)< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)
PT
0
26 tháng 3 2018
Đặt \(M=\frac{1}{4}+\frac{1}{16}+.....+\frac{1}{10000}\)
\(M=\frac{1}{2.2}+\frac{1}{4.4}+\frac{1}{6.6}+......+\frac{1}{100.100}\)
\(=\frac{1}{2.2}\left(1+\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{50.50}\right)\)
\(< \frac{1}{2.2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\right)\)
\(=\frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{4}.\left(1+1-\frac{1}{50}\right)< \frac{1}{4}.\left(1+1\right)=\frac{1}{4}.2=\frac{1}{2}\)
Vậy \(M< \frac{1}{2}\)
NT
1
JS
26 tháng 3 2020
Đặt: \(A=\frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}< \frac{1}{2}\)
Ta có: \(A=\frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}\)
\(\Rightarrow A=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)
\(\Rightarrow A< \frac{1}{4}.\frac{99}{50}\)
\(\Rightarrow A< \frac{99}{200}< \frac{1}{2}\)
Vậy: \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}< \frac{1}{2}\left(đpcm\right)\)
DM
1
8 tháng 2 2020
Đặt \(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)
\(A=\frac{1}{4}+\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)=\frac{1}{4}+\frac{1}{4}\cdot B\)
Ta có \(\frac{1}{2^2}< \frac{1}{1\cdot2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)
\(...\)
\(\frac{1}{50^2}< \frac{1}{49\cdot50}=\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{4}\cdot1=\frac{1}{2}\)
