Question

[vu thanh trung]

A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+\(\frac{1}{5.6}\)+....+\(\frac{1}{99.100}\)&B=\(\frac{2014}{51}\)+\(\frac{2018}{52}\)+\(\frac{2018}{53}\)+....+\(\frac{2018}{100}\)

Chung minh minh rang \(\frac{B}{A}\)la 1 so nguyen.

Ai làm đc mình dùng 2nick,mình tick cho

Answer 1

[๛Ňɠũ Vị Čáէツ]

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

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

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

   \(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)

   \(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(B=\frac{2018}{51}+\frac{2018}{52}+\frac{2018}{53}+...+\frac{2018}{100}\)

   \(=2018.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\)

\(\Rightarrow\frac{B}{A}=\frac{2018\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}\)

            \(=2018\)

Vậy \(\frac{B}{A}\)là 1 số nguyên

!!!