Question

[전정국]

cho A = 1/1*2+1/3*4+...+1/99*100 và B= 2015/51+2015/52+2015/53+...+2015/100. Chứng minh rằng B chia hết cho A

Answer 1

[Nguyễn Phạm Hồng Anh]

Ta có : \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

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

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

\(B=\frac{2015}{51}+\frac{2015}{52}+...+\frac{2015}{100}\)

    \(=2015\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)

\(\Rightarrow\) \(\frac{B}{A}=\frac{2015\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}=2015\)

\(\Rightarrow\) \(B⋮A\)