Question

Chứng minh rằng : \(A=1.2.3...2010\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)⋮2011\)

Giúp mình vs

Answer 1

Ta có: \(A=1.2.3...2010\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)\)

\(=\)1.2.3...2010\([\left(1+\frac{1}{2010}\right)+\left(\frac{1}{2}+\frac{1}{2009}\right)+...+\left(\frac{1}{1005}+\frac{1}{1006}\right)]\)

\(=\)\(1.2.3...2010\left(\frac{2011}{2010}+\frac{2011}{2009.2}+...+\frac{2011}{1005.1006}\right)\)

\(=2011\left(\frac{2010!}{2010}+\frac{2010!}{2009.2}+...+\frac{2010!}{1005.1006}\right)\)

Suy ra: A ⋮ 2011

Vậy A ⋮ 2011