chung minh rang 1\42 +1\62+1\82+ .......+1\(2.n)2<1\4
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S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)
(*)Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12
(*)Ta lại có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60
=>S<1/5+1/12.3+1/60.3
S<1/5+1/4+1/20
S<1/2
S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)
(*)Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12
(*)Ta lại có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60
=>S<1/5+1/12.3+1/60.3
S<1/5+1/4+1/20
S<1/2
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Lời giải:
Gọi vế trái là $A$
$2A=\frac{2}{2^2}+\frac{2}{4^2}+\frac{2}{6^2}+...+\frac{2}{2022^2}$
Xét số hạng tổng quát:
$\frac{2}{n^2}$. Ta sẽ cm $\frac{2}{n^2}< \frac{1}{(n-1)n}+\frac{1}{n(n+1)}(*)$
$\Leftrightarrow \frac{2}{n^2}< \frac{n+1+n-1}{n(n-1)(n+1)}$
$\Leftrightarrow \frac{2}{n^2}< \frac{2}{(n-1)(n+1)}$
$\Leftrightarrow \frac{2}{n^2}< \frac{2}{n^2-1}$ (luôn đúng)
Thay $n=2,4,...., 2022$ vào $(*)$ ta có:
$\frac{2}{2^2}< \frac{1}{1.2}+\frac{1}{2.3}$
$\frac{2}{4^2}< \frac{1}{3.4}+\frac{1}{4.5}$
.......
Suy ra: $2A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2022.2023}$
$2A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2022}-\frac{1}{2023}$
$2A< 1-\frac{1}{2023}< 1$
$\Rightarrow A< \frac{1}{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8
Dễ thấy: B=122+132+...+182B=122+132+...+182<A=11⋅2+12⋅3+...+17⋅8(1)<A=11⋅2+12⋅3+...+17⋅8(1)
Ta có:A=11⋅2+12⋅3+...+17⋅8A=11⋅2+12⋅3+...+17⋅8
=1−12+12−13+...+17−18=1−12+12−13+...+17−18
=1−18<1(2)=1−18<1(2)
Từ (1);(2)(1);(2) ta có: B<A<1⇒B<1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{8}=\frac{1}{8}\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}<\frac{3}{10}\)
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}<\frac{3}{40}\)
-> A <\(\frac{1}{8}+\frac{3}{10}+\frac{3}{40}=\frac{20}{40}=\frac{1}{2}\)
ta có:
\(\frac{1}{4^2}+\frac{1}{6^2}+..+\frac{1}{\left(2n\right)^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+...+\frac{1}{2^2.n^2}\)
\(=\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+..+\frac{1}{2^2}.\frac{1}{n^2}=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)=\frac{1}{4}.\left(\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{n^2}\right)\)
mà 1/2^2+1/3^2+..+1/n^2 < 1(cái này bn tự c/nm đc chứ?)
=>\(\frac{1}{4}.\left(\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{n^2}\right)<\frac{1}{4}\left(đpcm\right)\)
very sorry mik mới lớp 5 à nếu biết mik sẽ giải giùm bạn ! ^_^