Question

Tính:

\(\dfrac{1.2010+2.2009+3.2008+...+2010.1}{\left(1+2+3+...+2010\right)+\left(1+2+3+...+2009\right)+...\left(1+2\right)+1}\)

Answer 1

\(=\dfrac{1\cdot2010+2\cdot2009+3\cdot2008+...+2010\cdot1}{\left(1+1+...+1\right)+\left(2+2+...+2\right)+.....+\left(2009+2009\right)+2010}\\ =\dfrac{1\cdot2010+2\cdot2009+3\cdot2008+...+2010\cdot1}{1\cdot2010+2\cdot2009+3\cdot2008+...+2010\cdot1}\\ =1\)