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A=\(\left(1-\dfrac{1}{1+2}\right).\left(1-\dfrac{1}{1+2+3}\right).....\left(1-\dfrac{1}{1+2+3+4+...+2011}\right)\)

H24
3 tháng 3 2018 lúc 18:01

\(A=\left(1-\dfrac{1}{1+2}\right).\left(1-\dfrac{1}{1+2+3}\right)....\left(1-\dfrac{1}{1+2+...+2010}\right)\left(1-\dfrac{1}{1+2+...+2011}\right)\)\(A=A_1.A_2...A_n\) (n = [2,... 2011])

\(A_n=1-\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n.\left(n+1\right)-2}{n.\left(n+1\right)}=\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\) \(A_1=\dfrac{\left(2-1\right)\left(2+2\right)}{2\left(2+1\right)}=\dfrac{1.4}{2.3}\)

\(A_2=\dfrac{\left(3-1\right)\left(3+2\right)}{3\left(3+1\right)}=\dfrac{2.5}{3.4}\)

\(A_3=\dfrac{\left(4-1\right)\left(4+2\right)}{4\left(4+1\right)}=\dfrac{3.6}{4.5}\)

..

\(A=\dfrac{1.4.2.5.3.6.4.7...\left(2010\right).\left(2013\right)}{2.3.3.4.4.5...\left(2011\right)\left(2012\right)}=\dfrac{\left(1.2....2010\right)\left(4.5.6.2013\right)}{\left(2.3.4...2011\right)\left(3.4.5....2012\right)}\)

\(A=\dfrac{\left(1\right)\left(2013\right)}{\left(2011\right).\left(3\right)}=\dfrac{2013}{3.2011}=\dfrac{671}{2011}\)

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