CMR:
1/4+1/16+1/36+....+1/10000<1/2
K
Khách
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BC
2
27 tháng 4 2018
\(Đ\text{ặt }S=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}\)
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(S=\frac{1}{2^2}\cdot\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1\cdot2};\text{ }\frac{1}{3^2}< \frac{1}{2\cdot3};\text{ }...;\text{ }\frac{1}{50^2}< \frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)
\(\Rightarrow\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}\cdot2\)
\(\Rightarrow S< \frac{1}{2}\) (ĐPCM)
SK
30 tháng 4 2018
Đặt \(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{100^2}\)
\(\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)
\(\Rightarrow4A< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow4A=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow4A< 2-\frac{1}{50}< 2\)
\(\Rightarrow4A< 2\Rightarrow A< \frac{2}{4}=\frac{1}{2}\)
=>a<1/2
21 tháng 7 2017
Ta có: \(VT=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(4VT=\dfrac{1}{2^2:2^2}+\dfrac{1}{4^2:2^2}+\dfrac{1}{6^2:2^2}+...+\dfrac{1}{100^2:2^2}\)
\(4VT=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)
Lại có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(...\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\Rightarrow4VT-1< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)(*)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}\) (**)
Từ (*) và (**) \(\Rightarrow4VT< 2-\dfrac{1}{50}\)
\(\Rightarrow VT< \dfrac{1}{2}-\dfrac{1}{200}< VP\Rightarrow\) đpcm
b) Ta có: \(2VT=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)
\(2VT+VT=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)
\(3VT=1-\dfrac{1}{64}< 1\)
\(\Rightarrow VT< \dfrac{1}{3}\) (đpcm)
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 \(\frac{1}{2^2}< \frac{1}{1\cdot2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)
\(...\)
\(\frac{1}{50^2}< \frac{1}{49\cdot50}=\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{4}\cdot1=\frac{1}{2}\)
BN
0
13 tháng 10 2021
\(M=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{4^2}< \dfrac{1}{3\cdot4};...;\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)
\(\Rightarrow M< \dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-1-\dfrac{1}{2}-...-\dfrac{1}{50}\\ =\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\left(50.số\right)=\dfrac{50}{50}=1\)
Vậy \(M< 1\)
Mình chỉ so sánh với 1 được thôi à :((
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
\(\text{Đặt BT là A }\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(\text{Ta có:}\frac{1}{3^2}>\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\text{(để lại }\frac{1}{4}\text{ở đầu)}\)
\(\frac{1}{4^2}>\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)
.......
\(\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{101}\Rightarrow A=\frac{7}{12}-\frac{1}{101}=\frac{707-12}{1212}=\frac{695}{1212}>\frac{606}{1212}=\frac{1}{2}\)
\(\Rightarrow A>\frac{1}{2}\)
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 \(M< \frac{1}{2}\)
99/100
cho mk nhé