Câu hỏi của poison

Question

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mik can ngay baay h nen cac bn trl nhanh nha

Nguyễn Lê Phước Thịnh · Accepted Answer

$A=1\cdot2010+2\cdot2009+3\cdot2008+...+2010\cdot1$

$=1\left(2011-1\right)+2\left(2011-2\right)+3\left(2011-3\right)+...+2010\left(2011-2010\right)$

$=2011\left(1+2+3+...+2010\right)-\left(1^2+2^2+3^2+...+2010^2\right)$

$=2011\cdot\dfrac{2010\cdot2011}{2}-\dfrac{2010\left(2010+1\right)\left(2\cdot2010+1\right)}{6}$

$=2011\cdot2011\cdot1005-335\cdot2011\cdot4021$

$=2011\cdot335\left(2011\cdot3-4021\right)$

$=2011\cdot335\cdot2012$

$B=1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2010\right)$

$=1+\dfrac{2\cdot3}{2}+\dfrac{3\cdot4}{2}+...+\dfrac{2010\cdot2011}{2}$

$=\dfrac{1\cdot2+2\cdot3+...+2010\cdot2011}{2}$

$=\dfrac{1\left(1+1\right)+2\left(2+1\right)+...+2010\left(2010+1\right)}{2}$

$=\dfrac{\left(1^2+2^2+...+2010^2\right)+\left(1+2+...+2010\right)}{2}$

$=\dfrac{\dfrac{2010\left(2010+1\right)\left(2010\cdot2+1\right)}{6}+\dfrac{2010\cdot2011}{2}}{2}$

$=\dfrac{335\cdot2011\cdot4021+1005\cdot2011}{2}$

$=\dfrac{335\cdot2011\left(4021+3\right)}{2}$

$=335\cdot2011\cdot2012$

=>A/B=1

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