Tính B=$\frac{1}{3} \frac{1}{6}\left(1 2\right) \frac{1}{9}\left(1 2 3\right) ... \frac{1}{6045}\left(1 2 3 ... 2015\right)$13 16 (1 2) 19 (1 2 3) ... 16045 (1 2 3 ... 2015)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
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CMCT : ( tự CM )
Áp dụng bài toán trên ta có : \(B=\frac{1}{3}+\frac{1}{6}\vec{\left(\frac{\left(x+1\right).2}{2}\right)+\frac{1}{9}\left(\frac{\left(1+3\right).3}{2}\right)+...+}\frac{1}{6045}\left(\frac{\left(1+2015\right).2}{2}\right)\)
\(B=\frac{1}{3}+\frac{1}{2}+\frac{2}{3}+....+....\) ( tự lm tiếp )
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) áp dụng vào bài toán ta được
\(B=\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+...+2015\right)\)
\(=\frac{1}{3}+\frac{1}{2.3}.\frac{2.3}{2}+\frac{1}{3.3}.\frac{3.4}{2}+...+\frac{1}{2015.3}.\frac{2015.2016}{2}\)
\(=\frac{1}{3}\left(1+\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}\right)\)
\(=\frac{1}{6}\left(2+3+4+...+2016\right)=\frac{1}{6}.\frac{2015.2018}{2}=\frac{2033135}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
MK ghi sai để mk sửa lại nha