Tính
\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)....\left(1-\frac{1}{1+2+3+...+2006}\right)\)
K
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9 tháng 2 2019
\(A=(1-\frac{1}{1+2})(1-\frac{1}{1+2+3})(1-\frac{1}{1+2+3+4})...(1-\frac{1}{1+2+3+...+2006})\)
\(A=(1-\frac{1}{3})(1-\frac{1}{6})(1-\frac{1}{10})...(1-\frac{1}{2013021})\)
\(A=\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}....\frac{2013020}{2013021}\)
9 tháng 2 2019
Sorry bạn máy tính mình có chút vấn đề để mk làm tiếp :
\(A=\frac{4}{6}\cdot\frac{10}{12}\cdot\frac{18}{20}....\cdot\frac{4026040}{4026042}\)
\(A=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\frac{3\cdot6}{4\cdot5}\cdot...\cdot\frac{2005\cdot2008}{2006\cdot2007}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2005}{2\cdot3\cdot4\cdot...\cdot2006}\cdot\frac{4\cdot5\cdot6\cdot...\cdot2008}{3\cdot4\cdot5\cdot...\cdot2007}\)
\(A=\frac{1}{2006}\cdot\frac{2008}{3}=\frac{1004}{3009}\)
P/S : Hoq chắc :>
23 tháng 4 2017
\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)......\left(1-\frac{1}{1+2+3+...+2006}\right)\)
\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{2013021}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}....\frac{2013020}{2013021}\)
\(A=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}......\frac{4026040}{4026042}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}......\frac{2005.2008}{2006.2007}\)
\(A=\frac{1.2.3.....2005}{2.3.4....2006}.\frac{4.5.6....2008}{3.4.5...2007}\)
\(A=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)
17 tháng 6 2017
Ta thấy: \(1-\frac{1}{1+2+3+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\Rightarrow A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
\(=\left(\frac{\left(2-1\right)\left(2+2\right)}{2\left(2+1\right)}\right)\left(\frac{\left(3-1\right)\left(3+2\right)}{3\left(3+1\right)}\right)\left(\frac{\left(4-1\right)\left(4+2\right)}{4\left(4+1\right)}\right)...\left(\frac{\left(2006-1\right)\left(2006+2\right)}{2006\left(2006+1\right)}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{2005.2008}{2006.2007}=\frac{\left(1.2.3...2005\right)\left(4.5.6...2008\right)}{\left(2.3.4...2006\right)\left(3.4.5...2007\right)}\)
\(=\frac{1.2008}{2006.3}=\frac{1004}{1003.3}=\frac{1004}{3009}\)
Vậy \(A=\frac{1004}{3009}\)
15 tháng 1 2017
Ta có: \(1+2+..+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow\frac{1}{1+2+...+n}=\frac{2}{n\left(n+1\right)}\)
\(\Rightarrow1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Áp dụng vào bài toán ta được
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+2006}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.....\frac{2005.2008}{2006.2007}=\frac{1}{3}.\frac{2008}{2006}=\frac{1004}{3009}\)
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
\(A=\left(1-\frac{1}{\frac{\left(1+2\right).2}{2}}\right)\left(1-\frac{1}{\frac{\left(1+3\right).3}{2}}\right)...\left(1-\frac{1}{\frac{\left(1+2006\right).2006}{2}}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{2007.2006-2}{2006.2007}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}....\frac{2007.2006-2}{2006.2007}\) (1)
xét thấy:2007.2006-2=2006.(2008-1)+2006-2008=2006.(2008-1+1)-2008=2008.(2006-1)=2008.2005 (2)
(1),(2)\(=>A=\frac{4.1}{2.3}.\frac{5.2}{3.4}.\frac{6.3}{4.5}....\frac{2008.2005}{2006.2007}\)
\(A=\frac{\left(4.5.6...2008\right)\left(1.2.3...2005\right)}{\left(2.3.4....2006\right)\left(3.4.5...2007\right)}=\frac{2008}{2006.3}=\frac{1004}{3009}\)
Vậy A=1004/3009
dung hay sai zday