Question

\(Cho:\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{2017.2018}\)

\(B=\frac{1}{1010.2018}+\frac{1}{1011.2017}+\frac{1}{1012.2016}+.....+\frac{1}{2018.1010}\)

Chứng minh rằng: \(\frac{A}{B}\) là số nguyên

 

Answer 1

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}-1-\frac{1}{2}-...-\frac{1}{1009}\)

\(A=\frac{1}{1010}+\frac{1}{2000}+...+\frac{1}{2018}\)

\(B=3028.\left(\frac{1}{1010.2018}+...+\frac{1}{2018.1010}\right)\)

\(B=\frac{3028}{1010.2018}+...+\frac{3028}{2018.1010}\)

\(B=\frac{1}{1010}+\frac{1}{2018}+...+\frac{1}{2018}+\frac{1}{1010}\)

\(B=2.\left(\frac{1}{1010}+...+\frac{1}{2018}\right)\)

\(=>\frac{A}{B}=\frac{1}{2}\)

Answer 2

Linh Phương Ngô chứng minh a/b là số nguyên cơ mà

Answer 3

Thế thì mình k biết.