Question

[Vũ Thị Kim Oanh]

Cho: A = 1/1.2 + 1/3.4 + 1/5.6 + .... +1/99.100

chứng tỏ rằng:  7/12 < A < 5/6

Answer 1

[Nguyễn Thị BÍch Hậu]

\(A=\frac{1}{2}+\frac{1}{12}+...+\frac{1}{9900}>\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(1-\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)-\frac{1}{100}

Answer 2

[Vũ Anh Đức]

Ta có: 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+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100-2(1/2+1/4+1/6+...+1/100)

             =1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100-(1+1/2+1/3+1/4+...+1/50)

             =1/26+1/27+1/28+...+1/100)

Do đó A=(1/51+1/52+...+1/75)+(1/76+1/77+...+1/100)

Ta có 1/51>1/52>...>1/75 và 1/76>1/77>...>1/100 nên

A>1/75.25+1/100.25=1/3+1/4=7/12

A<1/51.25+1/76.25<1/50.25+1/75.25=1/2+1/3=5/6

Vậy nên 7/12<A<5/6