Question

thực hiện phép tính

\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...2012}\right)\)

Answer 1

Lời giải:
\(A=(1-\frac{1}{\frac{2(2+1)}{2}})(1-\frac{1}{\frac{3(3+1)}{2}})....(1-\frac{1}{\frac{2012(2012+1)}{2}})=(1-\frac{2}{6})(1-\frac{2}{12})(1-\frac{2}{20})....(1-\frac{2}{2012.2013})\)

Xét thừa số tổng quát:

\(1-\frac{2}{n(n+1)}=\frac{n^2+n-2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}\)

Do đó:

\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{2011.2014}{2012.2013}\\ =\frac{1.2.3...2011}{2.3.4...2012}.\frac{4.5.6...2014}{3.4.5...2013}\\ =\frac{1}{2012}.\frac{2014}{3}=\frac{1007}{3018}\)