Question

Cho A=\(\frac{1}{201}\)+\(\frac{1}{202}\)+\(\frac{1}{203}\)+...+\(\frac{1}{300}\).Chứng minh rằng A<\(\frac{9}{20}\)? Làm ơn giúp mik nha!

Answer 1

Ta có: A=1/201+1/202+1/203+...+1/300

=(1/201+1/202+...+1/250)+(1/251+1/252+...+1/300)

Ta có

1/201+1/202+...+1/250<1/200+1/200+...+1/200=50.1/200=50/200=1/4                   (1)

1/251+1/252+...+1/300<1/250+1/250+...+1/250=50.1/250=50/250=1/5                   (2)

từ (1) và (2)=> A<1/4+1/5=>A<9/20

Vậy A<9/20

~~~CHÚC BẠN HỌC GIỎI~~~        

=>A=

Answer 2

Ta có: A=1/201+1/202+1/203+...+1/300

=(1/201+1/202+...+1/250)+(1/251+1/252+...+1/300)

Ta có

1/201+1/202+...+1/250<1/200+1/200+...+1/200=50.1/200=50/200=1/4                   (1)

1/251+1/252+...+1/300<1/250+1/250+...+1/250=50.1/250=50/250=1/5                   (2)

từ (1) và (2)=> A<1/4+1/5=>A<9/20

Vậy A<9/20