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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 \(\frac{1}{2^2}< \frac{1}{1\cdot2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)
\(...\)
\(\frac{1}{50^2}< \frac{1}{49\cdot50}=\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{4}\cdot1=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/4 = 1/(2*2) < 1/(1*2) = 1/2 - 1/4
tương tự ta có
1/16 < 1/(2*4) = 1/4 - 1/8
1/36 < 1/(4*6) = 1/8 - 1/12
1/64 < 1/(6*8) = 1/12 - 1/16
1/100 < 1/(8*10) = 1/16 - 1/20
1/144 < 1/(10*12) = 1/20 - 1/24
1/196 < 1/(12* 14) = 1/24 - 1/28
cộng hết lại
=> 1/4 + 1/16 + ......+ 1/100 + 1/144 + 1/196 < 1/2 - 1/28 < 1/2 => đpcm
ta có
1/4 = 1/(2*2) < 1/(1*2) = 1/2 - 1/4
tương tự ta có
1/16 < 1/(2*4) = 1/4 - 1/8
1/36 < 1/(4*6) = 1/8 - 1/12
1/64 < 1/(6*8) = 1/12 - 1/16
1/100 < 1/(8*10) = 1/16 - 1/20
1/144 < 1/(10*12) = 1/20 - 1/24
1/196 < 1/(12* 14) = 1/24 - 1/28
cộng hết lại
=> 1/4 + 1/16 + ......+ 1/100 + 1/144 + 1/196 < 1/2 - 1/28 < 1/2 => đpcm
Tick đúng nha bạn
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A=1/22+1/32+...+1/92
Ta có:1/22>1/2.3,1/32>1/3.4,...,1/92>1/9.10
⇒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>2/5(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}< \frac{1}{2}\left(đpcm\right)\)