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Tổng các số tự nhiên từ 1 đến n là \(\frac{n\left(n+1\right)}{2}\)
Do đó \(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2011}.\frac{2011.2012}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2012}{2}\)
\(=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2012}{2}\right)-\frac{1}{2}\)
\(=\frac{1+2+3+...+2012}{2}-\frac{1}{2}\)
\(=\frac{\frac{2012.2013}{2}}{2}-\frac{1}{2}\)
\(=1012538,5\)
Vậy ....
\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)......\left(1-\frac{1}{2011}\right)\)
\(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2010}{2011}\)
\(=\frac{6.7.8.9.....2010}{7.8.9.10.....2011}\)
\(=\frac{6}{2011}\)
=\(\frac{2}{1+2}.\frac{2+3}{1+2+3}.\frac{2+3+4}{1+2+3+4}...\frac{2+3+4+...+2011}{1+2+3+....+2011}\)
=\(\frac{2}{\frac{\left(2+1\right).2}{2}}.\frac{\left(2+3\right).2}{\frac{2}{\frac{\left(3+1\right).3}{2}}}....\frac{\left(2+2011\right)\left(2011-1\right)}{\frac{2}{\frac{\left(2011+1\right)2011}{2}}}\)
=\(\frac{4}{\left(2+1\right).2}\frac{\left(2+3\right).2}{\left(3+1\right).3}....\frac{(2+2011)\left(2011-1\right)}{\left(2011+1\right)2011}\)
=\(\frac{\left(1.4\right)\left(5.2\right)....\left(2013.2010\right)}{\left(3.2\right).\left(4.3\right)....\left(2012.2011\right)}\)
=\(\frac{\left(1.2.3...2010\right)\left(4.5.6...2013\right)}{\left(2.3.4...2011\right)\left(3.4.5....2012\right)}\)
=\(\frac{1}{2011}.\frac{2013}{3}\)=\(\frac{671}{2011}\)
Mk nghĩ vậy. Chắc là đúng đấy