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Ta có:
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{25}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(=\frac{1}{25}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{25}+\frac{1}{6}-\frac{1}{101}>\frac{1}{6}+\frac{1}{25}-\frac{1}{100}=\frac{1}{6}+\frac{3}{100}>\frac{1}{6}\left(1\right)\)
\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\left(2\right)\)
Từ (1) và (2) suy ra:\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(đpcm\right)\)
đạt 1/52+.........+1/1002=S
1/52>1/5*6
.....................
1/1002>1/100*101
=>S>1/5*6+.............+1/100*101=1/5-1/6+....+1/100-1/101=1/5-1/101=96/505>96/576=1/6
vậ S>1/6
1/52<1/4*5
.....................
1/1002<1/99*100
=>S<1/4*5+................+1/99*100=1/4-1/5+.....+1/99-1/100=1/4-1/100=6/25<6/24=1/4
Vậy 1/6<S<1/4
- Đặt \(S=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2008}{3^{2008}}\)(1)
- Ta có: \(\frac{1}{3}S=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{2007}{3^{2008}}+\frac{2008}{3^{2009}}\)(2)
- Trừ vế với vế 2 đửng thức (1) và (2) ta có:
\(S-\frac{1}{3}S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}-\frac{2008}{3^{2009}}<\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)(3)
- Đặt \(P=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
- \(\left(1-\frac{1}{3}\right)P=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}-\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2008}}+\frac{1}{3^{2009}}\right)=\frac{1}{3}-\frac{1}{3^{2009}}<\frac{1}{3}\)
- \(\frac{2}{3}P<\frac{1}{3}\Rightarrow P<\frac{1}{2}\)(4)
- Từ (3) và (4)
\(\Rightarrow\frac{2}{3}S<\frac{1}{2}\Rightarrow S<\frac{3}{4}\)(ĐPCM)
\(P=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1+\frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(P< 1+\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}=\frac{7}{4}-\frac{1}{2019}< \frac{7}{4}\)
Đặt \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}=A\)
ta có :\(\frac{1}{2^2}=\frac{1}{2\cdot2}=\frac{1}{4}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(...\)
\(\frac{1}{1990^2}=\frac{1}{1990\cdot1990}< \frac{1}{1989\cdot1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2\cdot3}+...+\frac{1}{1989\cdot1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\left(ĐPCM\right)\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
hk tốt #
Ta có \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{1990^2}< \frac{1}{1989.1990}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)
\(< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)
\(\Rightarrow\)Bài toán được chứng minh
Đăng từ bài thôi bạn à!
a) Áp dụng công thức: \(\frac{1}{a-1}-\frac{1}{a}=\frac{1}{\left(a-1\right)a}>\frac{1}{a.a}=\frac{1}{a^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4^2}< \frac{1}{3}-\frac{1}{4}\)
..............................
\(\frac{1}{n^2}< \frac{1}{n-1}-\frac{1}{n}\)
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\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}=\frac{1}{n+1}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\) (đpcm)
=>1/2^2+...+1/1990^2<1/1990<3/4