Question

Tính B=\(\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+3+...+2015\right)\)

Answer 1

Ta có: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) áp dụng vào bài toán ta được

\(B=\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+...+2015\right)\)

\(=\frac{1}{3}+\frac{1}{2.3}.\frac{2.3}{2}+\frac{1}{3.3}.\frac{3.4}{2}+...+\frac{1}{2015.3}.\frac{2015.2016}{2}\)

\(=\frac{1}{3}\left(1+\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}\right)\)

\(=\frac{1}{6}\left(2+3+4+...+2016\right)=\frac{1}{6}.\frac{2015.2018}{2}=\frac{2033135}{6}\)