Question

[Hoàng Thị Trà My]

cho A=1/1.2+1/3.4+1/5.6+.....+1/101+102

B=1/52.102+1/53.101+1/54.100+.....+1/101.53+102.52

Chứng minh rằng :A/B là số nguyên

Answer 1

[Y]

+ \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{101}-\frac{1}{102}\)

\(A=\left(1+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{102}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{102}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{102}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{102}\right)-\left(1+\frac{1}{2}+...+\frac{1}{51}\right)\)

\(A=\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{102}\)

+ \(154B=\frac{52+102}{52\cdot102}+\frac{53+101}{53\cdot101}+...+\frac{102+52}{102\cdot52}\)

\(154B=\frac{1}{52}+\frac{1}{102}+\frac{1}{53}+\frac{1}{101}+...+\frac{1}{101}+\frac{1}{53}+\frac{1}{102}+\frac{1}{52}\)

\(154B=2\left(\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\right)\)

\(B=\frac{1}{77}\left(\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\right)\)

Do đó : \(\frac{A}{B}=\frac{1}{\frac{1}{77}}=77\) là số nguyên

Answer 2

[Hoàng Thị Trà My]

thank you bạn Y nha

kết bạn vs mình ik