B = 1/3+1/6(1+2)+1/9(1+2+3)+...+1/6045(1+2+3+...+2015)
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)
A=[(3/11+1-3/7)/(3+9/11-9/7)]-[(1/3+0.25-1/5+0.125)/(7/6+7/8-0.7+7/16)]
B=1/3+1/6*(1+2)+1/9(1+2+3)+...+1/6045*(1+2+3+..+2015)
giúp vs cần gấp
Tính B=\(\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+3+...+2015\right)\)
Ta có: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) áp dụng vào bài toán ta được
\(B=\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+...+2015\right)\)
\(=\frac{1}{3}+\frac{1}{2.3}.\frac{2.3}{2}+\frac{1}{3.3}.\frac{3.4}{2}+...+\frac{1}{2015.3}.\frac{2015.2016}{2}\)
\(=\frac{1}{3}\left(1+\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}\right)\)
\(=\frac{1}{6}\left(2+3+4+...+2016\right)=\frac{1}{6}.\frac{2015.2018}{2}=\frac{2033135}{6}\)
B=\(\frac{1}{3}+\frac{1}{6}\cdot\left(1+2\right)+\frac{1}{9}\cdot\left(1+2+3\right)+...+\frac{1}{6045}\cdot\left(1+2+3+...+2015\right)\)
Tính \(B=\frac{1}{3}+\frac{1}{6}.\left(1+2\right)+\frac{1}{9}.\left(1+2+3\right)+...+\frac{1}{6045}.\left(1+2+..+2015\right)\)
Tính B=\(\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+3+...+2015\right)\)
Tính B=\(\frac{1}{3}+\frac{1}{6}.\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+....+\frac{1}{6045}\left(1+2+3+\text{4}+.....+2015\right)\)
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 )
Tính A=\(\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+...+2015\right)\)
\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
tính \(A=\dfrac{1}{3}+\dfrac{1}{6}\left(1+2\right)+\dfrac{1}{9}\left(1+2+3\right)+...+\dfrac{1}{6045}\left(1+2+3+...+2015\right)\)