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ta có 1/51>1/100
1/52>1/100
..................
1/100=1/100
\(\Rightarrow\)S=1/51+1/52+...+1/100>(1/100+1/100+...+1/100)=1/100.50=1/2
\(\Rightarrow\)S>\(\frac{1}{2}\)
cái chỗ 1/100+1/100+...+1/100 có 50 số bạn nhá
chúc bạn học tốt~
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+............+\frac{1}{100}\)
\(=\left(\frac{1}{51}+\frac{1}{52}+.........+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+........+\frac{1}{100}\right)\)
\(>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{1}{2}\)
\(\left(\frac{1}{51}+\frac{1}{52}+..........+\frac{1}{75}\right)+\left(\frac{1}{76}+........+\frac{1}{100}\right)\)
\(< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}< 1\)
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(H=\frac{1}{51}+\frac{1}{52}+\frac{1}{52}+....+\frac{1}{100}\)
\(\Rightarrow H>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}\)
\(\Rightarrow H>\frac{1}{100}.50\)
\(\Rightarrow H>\frac{1}{2}\)
Lại có :
\(H=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+.....+\frac{1}{100}\)
\(\Rightarrow H< \frac{1}{51}+\frac{1}{51}+\frac{1}{51}+........+\frac{1}{51}\)
\(\Rightarrow H< \frac{1}{51}.50\)
\(\Rightarrow H< \frac{50}{51}\)
\(\Rightarrow H< 1\)
Vậy \(\frac{1}{2}< H< 1\left(ĐPCM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt A = 1/1*2 + 1/3*4 + 1/5*6 + ... + 1/99*100
= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/99 - 1/100
= (1 + 1/3 + 1/5 + ... + 1/99) - (1/2 + 1/4 + 1/6 + ... + 1/100)
= 1 + 1/2 + 1/3 + ... + 1/100 - 2(1/2 + 1/4 + 1/6 + .... + 1/100)
= 1 + 1/2 + 1/3 + ... + 1/100 - 1 - 1/2 - 13 - ... - 1/50
= 1/51 + 1/52 + 1/53 + ... + 1/100
thay vào ra E = 1
Biến đổi mẫu ta được:
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\Rightarrow E=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}=1\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=\left(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{75}\right)+...+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)
\(\frac{1}{51}+\frac{1}{52}+..+\frac{1}{75}< \frac{1}{2}\)
\(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{75}+\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=\frac{1}{3}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{2}+\frac{1}{3}=\frac{5}{6}< 1\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{3}\)
\(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)
\(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{4}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)