tính A = 1/3+1/6*(1+2)+...+1/6045*(1+2+...+2015)
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\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
\(B=\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+....+\frac{1}{6045}\left(1+2+3+...+2015\right)\)
\(=\frac{1}{3}+\frac{1}{6}.\frac{2.3}{2}+\frac{1}{9}.\frac{3.4}{2}+....+\frac{1}{6045}.\frac{2015.2016}{2}\)
\(=\frac{1}{3}\left(1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2015}.\frac{2015.2016}{2}\right)\)
\(=\frac{1}{3}\left(\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.....+\frac{2016}{2}\right)\)
\(=\frac{1}{3}\left(\frac{2+3+4+...+2016}{2}\right)\)
\(=\frac{1}{3}.\frac{2016.2017-1}{2}\)
\(=\frac{1}{3}.\frac{4066271}{2}=\frac{4066271}{6}\)
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 )