Question

Chứng minh rằng

\(A=1\cdot2\cdot3\cdot...\cdot2010\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2010}\right)⋮2011\)

Answer 1

\(\dfrac{1}{2}\)\(\dfrac{1}{3}\)+\(\dfrac{1}{2010}\)

\(\dfrac{1}{2010}\))+(\(\dfrac{1}{2}\)+\(\dfrac{1}{2009}\))+(\(\dfrac{1}{3}\)+\(\dfrac{1}{2008}\))+...+(\(\dfrac{1}{1005}\)+\(\dfrac{1}{1006}\))]

= \(\dfrac{2011}{2010}\)+\(\dfrac{2011}{2009\cdot2}\)+\(\dfrac{2011}{2008\cdot3}\)++...+\(\dfrac{2011}{1006\cdot1005}\))

= 2011*\(\dfrac{2010!}{2010}\)+\(\dfrac{2010!}{2009\cdot2}\)+\(\dfrac{2010!}{2008\cdot3}\)++...+\(\dfrac{2010!}{1006\cdot1005}\))

Answer 2

Có: \(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2010}\)\(=\dfrac{\left(\dfrac{1}{1}+\dfrac{1}{2010}\right).2010}{2}\)\(=\dfrac{2011}{2}\)

\(\Rightarrow A=1\cdot2\cdot3\cdot...\cdot2010\cdot\dfrac{2011}{2}\)

=\(1\cdot3\cdot4\cdot...\cdot2010.2011⋮2011\)