`Answer:`

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

\(=1+\frac{1}{2}.3+...+\frac{1}{2}.210\)

\(=1+1,5+2+...+10,5\)

\(=\frac{\left(10,5+1\right)[\left(10,5-1\right):0,5+1]}{2}\)

\(=\frac{230}{2}\)

\(=115\)

Ta có:

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

\(=1+\dfrac{1}{2}.\dfrac{2\left(2+1\right)}{2}+\dfrac{1}{3}.\dfrac{3\left(3+1\right)}{2}+...+\dfrac{1}{20}.\dfrac{20\left(20+1\right)}{2}\)

\(=\dfrac{2}{2}+\dfrac{2+1}{2}+\dfrac{3+1}{2}+...+\dfrac{20+1}{2}\)

\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{20}{2}\)

\(=\dfrac{2+3+4+...+20}{2}=\dfrac{\dfrac{20\left(20+1\right)}{2}-1}{2}\)

\(=\dfrac{209}{2}\)

Vậy \(B=\dfrac{209}{2}\)

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

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

\(=\frac{2}{2}+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{20+1}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{20}{2}\)

\(=\frac{2+3+4+...+20}{2}=\frac{\frac{20\left(20+1\right)}{2}-1}{2}=\frac{209}{2}\)

Đinh Đức Hùng trả lời sai rồi

ở dòng thứ 4 của bài làm của Đinh Đức Hùng thì phân số 20/2 phải là 21/2 mới đúng.