Question

Chứng tỏ rằng : \(\frac{1}{101}\)+ \(\frac{1}{102}\)+ \(\frac{1}{103}\)+ .......+ \(\frac{1}{200}\)>\(\frac{1}{2}\)

Answer 1

1/2=1/200+1/200+1/200+.....+1/200 (có 100 số )

1/101+1/102+....+1/200(có 100 số )

Vì 1/101>1/200

1/102>1/100

......

1/199>1/200

1/200=1/200

=>1/101+1/102+.....+1/200>1/200+1/200+...+1/200 có 100 số

=>1/101+1/102+.....+1/200>1/2

Answer 2

Ta thấy \(\frac{1}{101}>\frac{1}{200};\frac{1}{102}>\frac{1}{200};\frac{1}{103}>\frac{1}{200};....;\frac{1}{200}=\frac{1}{200}\)

Mà dãy \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{200}\)có 100 phân số nên : 

\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)( có 100 phân số \(\frac{1}{200}\))

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{200}.100=\frac{1.}{2}\left(đpcm\right)\)

Answer 3

tổng \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)                có 100 số hạng

Ma cac so hang cua tong trên lớn hơn 1/200 = > tong tren lon hon 1/200 * 100 = 1/2 (dpcm)