Question

Chứng minh: \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}\right)\)chia hết cho 11

Answer 1

Ta có: \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}\right)=\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)
      \(=\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}=99.\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right)\)
      \(=9.11\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right)\)
Vậy: đpcm