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Ta có : \(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4.4}< \frac{1}{3.4}\)
...................
\(\frac{1}{100.100}< \frac{1}{99.100}\)
Suy Ra : \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+......+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)
\(\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{100.100}< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Ta có : \(\frac{1}{2.2}\)\(< \frac{1}{1.2}\)
\(\frac{1}{3.3}\)\(< \frac{1}{2.3}\)
\(\frac{1}{4.4}\)\(< \frac{1}{3.4}\)
...... .... ......
\(\frac{1}{100.100}\)\(< \frac{1}{99.100}\)
\(\Rightarrow\)\(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ \(\frac{1}{4.4}\)+ ..... + \(\frac{1}{100.100}\)< \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ..... + \(\frac{1}{99.100}\)
\(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ .... + \(\frac{1}{100.100}\)< \(1-\frac{1}{100}=\frac{99}{100}< 1\)
Bạn sai đè thì phải,đúng phải là 1/99
Ta thấy:Từ 1->1/100 có 100 số.
Ta có:100=1.100
Vì 1=1 ;1/2<1 ;1/3<1 ;1/4<1 ;... ;1/90<1 ;1/100<1.
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}< 1.100=100\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}< 100\)
Ta có:\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}< 1\)
=>A<1
1/2.2 < 1/1.2
1/3.3 < 1/2.3
..................
1/100.100 < 1/99.100
=> <
Ta có: \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\)
Vì \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
\(\frac{1}{4^2}<\frac{1}{3.4}\)
.....
\(\frac{1}{100^2}<\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}<1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<1\left(đpcm\right)\)
S= 1/2 - 1/2 + 1/3 - 1/3 + 1/4 - 1/4 +...+ 1/50 - 1/50
S= 0 + 0 + 0 +...+ 0
S= 0
chung to :C = \(\frac{1}{1.1!}+\frac{1}{2.2!}+\frac{1}{3.3!}+...+\frac{1}{2019.2019!}< \frac{3}{2}\)
Thấy : \(\frac{1}{1.1!}=\frac{1}{1}\)
\(\frac{1}{2.2!}=\frac{1}{4}\)
\(\frac{1}{3.3!}< \frac{1}{1.2.3}\)( Vì 3.3! > 1.2.3 )
...
\(\frac{1}{2019.2019!}< \frac{1}{2017.2018.2019}\)( vì 2019.2019! < 2017.2018.2019)
Cộng từng vế có :
\(\frac{1}{3.3!}+\frac{1}{4.4!}+...+\frac{1}{2019.2019!}< \frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)
\(\Rightarrow\frac{1}{1.1!}+\frac{1}{2.2!}+...+\frac{1}{2019.2019!}< \frac{1}{1}+\frac{1}{4}+\frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)
\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right):2\)
\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2018.2019}\right):2\)
\(\Rightarrow C< \frac{3}{2}-\frac{1}{2.2018.2019}\)
Vì \(\frac{1}{2.2018.2019}>0\Rightarrow C< \frac{3}{2}\)
Có : 1/2^2+1/3^2+....+1/100^2 < 1/1.2+1/2.3+....+1/99.100 = 1-1/2+1/2-1/3+....+1/99-1/100 = 1-1/100 < 1
=> ĐPCM
k mk nha
Đặt \(A=\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{100.100}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}\)
\(\Rightarrow A< \frac{99}{100}\)
Mà \(\frac{99}{100}< 1\Rightarrow A< \frac{99}{100}< 1\)
\(\Rightarrow A< 1\)