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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
xét vế trái
=(1+1/3+1/5+...+1/1989)-(1/2+1/4+...+1/1990)
=(1+1/2+1/3+1/4+...+1/1990)-2.(1/2+1/4+...+1/1990)
=(1+1/2+1/3+1/4+...+1/1990)-!1+1/2+1/3+1/4+...+1/995)
=1/996+1/997+.../1+1990
vậy 1-1/2+1/3-1/4+...-1/1990=1/996+1/997+...+1/1990
cmr 1-$\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......-\frac{1}{1990}=\frac{1}{996}+\frac{1}{997}+\frac{1}{998}+.......+\frac{1}{1990}$
- 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)
ta có biêu thức trên\(\: < \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\)=\(\frac{2012}{2013}< 1\)
do dó biểu thức <1
Đặt A = 1-1/2+1/3-1/4 +...+1/1989-1/1990
A= (1+1/3+1/5 +...+1/1989)- ( 1/2 + 1/4 +....+1/1990 )
A=(1+1/3+1/5 +...+1/1989) - 2(1/2+1/4+1/6+.....+1/1990)
A= (1+1/3+1/5 +...+1/1989)- (1+1/2+1/3+1/4 +...+1/995)
A= 1/996+1/997 +.....+1/1990 =VP (đpcm)
Chúc các bạn thành công :)
Có điều gì sai các bạn bẩu mình nha :)
A=
\(P=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1+\frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(P< 1+\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}=\frac{7}{4}-\frac{1}{2019}< \frac{7}{4}\)
=>1/2^2+...+1/1990^2<1/1990<3/4