Question

Chứng minh rằng :

\(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}<1\)

Answer 1

lop 6 phai ko

Answer 2

Có cần tính ra không?

Answer 3

Gọi A=1/101+...+1/200

=>A có số thừa số là (200-101):1+1=100 (thừa số)

=>1/101+...+1/200 <1/100+1/100+...+1/100 (100 ts 1/100)

=>1/101+...+1/200 <1(đpcm)

Answer 4

Ta có : \(\frac{1}{101}<\frac{1}{100};\frac{1}{102}<\frac{1}{100};...;\frac{1}{200}<\frac{1}{100}\)

=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{199}+\frac{1}{200}<\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{100}{100}=1\)

=> \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}+\frac{1}{200}<1\)

Answer 5

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{199}+\frac{1}{200}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+.........+\frac{1}{199}+\frac{1}{200}<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...............+\frac{1}{199}+\frac{1}{200}<1-\frac{1}{100}=\frac{99}{100}<1\)

\(\Leftrightarrow\frac{1}{101}+\frac{1}{102}+...........+\frac{1}{199}+\frac{1}{200}<1\) (ĐPCM)

Answer 6

\(\frac{1}{102}<\frac{1}{101};\frac{1}{102}<\frac{1}{101};..........\)

\(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}<\frac{1}{101}+\frac{1}{101}+....+\frac{1}{101}=\frac{1}{101}.100=\frac{100}{101}<1\)

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