Question

[Nguyễn Thị Thu Trang]

Tính B biết :

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)

Answer 1

[Nguyễn Ngọc Quý]

\(B=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+.....+\frac{1}{20}.\frac{20.21}{2}\)

\(B=\frac{2}{2}+\frac{2.3:2}{2}+\frac{3.4:3}{2}+.....+\frac{20.21:20}{2}\)

\(B=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+....+\frac{21}{2}\)

\(B=\frac{2+3+.....+21}{2}\)

\(B=\frac{\left(2+21\right).20}{2}:2=\frac{230}{2}:2=165:2=\frac{165}{2}\)