Câu hỏi của Luân Đào

Question

Chứng minh rằng

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

Anonymous · Accepted Answer

$A=1\cdot2\cdot3\cdot...\cdot2010(1+$ $\dfrac{1}{2}$ $+$ $\dfrac{1}{3}$+ $...+$ $\dfrac{1}{2010}$ $)$

$=$ $1\cdot2\cdot3\cdot...\cdot2010[(1+$ $\dfrac{1}{2010}$)+($\dfrac{1}{2}$+$\dfrac{1}{2009}$)+($\dfrac{1}{3}$+$\dfrac{1}{2008}$)+...+($\dfrac{1}{1005}$+$\dfrac{1}{1006}$)]

= $1\cdot2\cdot3\cdot...\cdot2010($ $\dfrac{2011}{2010}$+$\dfrac{2011}{2009\cdot2}$+$\dfrac{2011}{2008\cdot3}$++...+$\dfrac{2011}{1006\cdot1005}$)

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

$=> A⋮2011 (dpcm)$

Câu trả lời:

$A=1\cdot2\cdot3\cdot...\cdot2010(1+$ $\dfrac{1}{2}$ $+$ $\dfrac{1}{3}$+ $...+$ $\dfrac{1}{2010}$ $)$

$=$ $1\cdot2\cdot3\cdot...\cdot2010[(1+$ $\dfrac{1}{2010}$)+($\dfrac{1}{2}$+$\dfrac{1}{2009}$)+($\dfrac{1}{3}$+$\dfrac{1}{2008}$)+...+($\dfrac{1}{1005}$+$\dfrac{1}{1006}$)]

= $1\cdot2\cdot3\cdot...\cdot2010($ $\dfrac{2011}{2010}$+$\dfrac{2011}{2009\cdot2}$+$\dfrac{2011}{2008\cdot3}$++...+$\dfrac{2011}{1006\cdot1005}$)

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

$=> A⋮2011 (dpcm)$

Nguyen · Answer

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

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

=$1\cdot3\cdot4\cdot...\cdot2010.2011⋮2011$

Câu trả lời:

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

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

=$1\cdot3\cdot4\cdot...\cdot2010.2011⋮2011$

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