Question

giúp mình nhanh dùm

tính a=1+1/2.(1+2)+1/3.(1+2+3)+...+1/2018.(1+2+3+...+2018)

Answer 1

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.....+\frac{1}{2018}\left(1+2+3+...+2018\right)\)

\(=1+\frac{1}{2}\cdot\frac{2.\left(2+1\right)}{2}+\frac{1}{3}\cdot\frac{3.\left(3+1\right)}{2}+...+\frac{1}{2018}\cdot\frac{2018\left(2018+1\right)}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{2019}{2}\)

\(=\frac{2+3+4+...+2019}{2}\)

\(=\frac{\frac{2019\left(2019+1\right)}{2}-1}{2}=1019594.5\)