Question

Chứng minh rằng số tự nhiên A chia hết cho 101 với: 

A=1.2.3...99.100,(1\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}\))

Answer 1

Ta có: A=1.2.3.....99.100.(\(1+\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{99}+\dfrac{1}{100}\))

      \(=1.2.3...100\left[\left(1+\dfrac{1}{100}\right)+\left(\dfrac{1}{2}+\dfrac{1}{99}\right)+......+\left(\dfrac{1}{50}+\dfrac{1}{51}\right)\right]\)

      =>A= 1.2...100.\(\left[\dfrac{101}{100}+\dfrac{101}{2.99}+......+\dfrac{101}{50.51}\right]\)

       =1.2.....100.101\(\left[\dfrac{1}{100}+\dfrac{1}{2.99}+.....+\dfrac{1}{50.51}\right]⋮101\)

               Vậy A chia hết cho 101