cho A=\(\frac{1}{1^2}+\frac{1}{2^2}+......+\frac{1}{50^2}\)Chưng minh rang a<2
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
mỗi p/số của A đều bé hơn 1/1.2+1/2.3+1/3.4+......+1/49.50
A<1-1/2+1/2-1/3+1/3-1/4+..........+1/49-1/50(tách ra thành hiệu)
A<1-1/50
mà 1/50>0=>1-1/50<1<2
A<1-1/50<1<2
A<2
chúc học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A<\frac{1}{1\cdot2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}<1<2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/2.2 + 1/3.3 +......+ 1/50.50
A < 1/1.2 + 1/2.3 +......+ 1/49.50
A < 1 - 1/2 + 1/2 - 1/3 +.....+ 1/49 - 1/50
A < 1 - 1/50
A < 49/50 < 3/4
=> A < 3/4 (đpcm)
Hình như bạn Killua giải sai thì phải.. 49/50 > 3/4 chứ
Theo mình thì bài này nên giữ nguyên phân số 1/2^2( vì nó bằng 1/4)
Xét : B = 1/3^2 + 1/4^2 +...+ 1/50^2
=> B < 1/2.3 + 1/3.4 +...+ 1/49.50
=> B< 1/2-1/3+1/3-1/4+...+1/49-1/50
=> B < 1/2-1/50 < 1/2
Suy ra A < 1/2^2 + 1/2 = 3/4
Vậy A< 3/4
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(T=3\cdot5\cdot7\cdot.....\cdot49\)
\(\Rightarrow A\cdot T=\frac{T}{2}+\frac{T}{3}+\frac{T}{4}+....+\frac{T}{50}\)
\(2^4\cdot B\cdot T=\frac{2^4T}{2}+\frac{2^4T}{3}+\frac{2^4T}{4}+....+\frac{2^4T}{50}\left(1\right)\)
Tất cả các số hạng của (1) đều là stn ngoại trừ \(\frac{2^4T}{5}\)
\(\Rightarrow VP\notinℕ\Rightarrow VT\notinℕ\)
Mà \(2^4\inℕ\Rightarrow T\inℕ\)
\(\Rightarrow A\notinℕ\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Rightarrow2+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\Rightarrow\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=2\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
\(\Rightarrow\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}=1\Rightarrow\frac{a+b+c}{abc}=1\Rightarrow a+b+c=abc\)
\(\Rightarrowđpcm\)
Ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{2}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Rightarrow2^2=2+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Leftrightarrow2=.2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)
\(\Leftrightarrow\frac{a}{abc}+\frac{a}{abc}+\frac{b}{abc}=\frac{abc}{abc}\)
\(\Leftrightarrow a+b+c=abc\)
\(\RightarrowĐPCM\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge a+b+c+\frac{9}{a+b+c}\)
Đặt: \(a+b+c=t\le\frac{3}{2}\Leftrightarrow2t\le3\)
Ta có: Cần cm: \(t+\frac{9}{t}\ge\frac{15}{2}\Leftrightarrow\frac{t^2+9}{t}\ge\frac{15}{2}\Leftrightarrow2t^2+18-15t\ge0\)
\(\Leftrightarrow\left(2t^2-3t\right)+\left(18-12t\right)\ge0\Leftrightarrow t\left(2t-3\right)-6\left(2t-3\right)\ge0\Leftrightarrow\left(t-6\right)\left(2t-3\right)\ge0\)(đúng với \(t\le\frac{3}{2}\))
Dấu "=" khi \(a=b=c=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(.......\)
\(\frac{1}{50^2}< \frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Mà \(\frac{49}{50}< 2\)
\(\Rightarrow A< 2\)
CHO \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}.\)CHỨNG MINH A<2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2^2}< \frac{1}{1.2}\)
...................\(\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow A< 1-\frac{1}{50}< \frac{49}{50}< 1< 2\)
1/2^2<1/1*2;1/3^2<1/2*3;1/4^2<1/3*4;1/50^2<1/49*50
ta có:
=> 1/1^2+1/2*3+1/3*4+...+1/49*50
<=> 1/1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
<=> 1-1/50 < 2
=> A < 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{49.50}\)
\(A< 1+\frac{49}{50}\)
\(A< 1\frac{49}{50}\)
Mà \(\frac{49}{50}< 2\)nên A<2
A=1+[\(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\)
ta có \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{50^2}<\frac{1}{49.50}\)
=>A<1+\(\left[\frac{1}{1.2}+.........+\frac{1}{49.50}\right]\)
=>A<1+\(\left[\frac{1}{1}-\frac{1}{50}\right]\)
=>A<1+\(\frac{49}{50}\)
=>A<\(\frac{99}{50}\) <2
=>A<2
K MÌNH NHA BÀI NÀY MÌNH GHI MỎI TAY LẮM
A=\(\frac{1}{1^2}+\frac{1}{2^2}+....+\frac{1}{50^2}\)
A<\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49\cdot50}\)
A<1+\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
A<1+\(\left(1-\frac{1}{50}\right)\)
A<1+\(\frac{49}{50}\)
=>A<2