Question

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

Answer 1

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

\(\Rightarrow B=1+\dfrac{1}{2}.2.3\div2+\dfrac{1}{3}.3.4\div2+...+\dfrac{1}{20}.20.21\div2\)

\(\Rightarrow B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\)

\(\Rightarrow B=\dfrac{2+3+4+...+21}{2}\)

\(\Rightarrow B=\dfrac{230}{2}\)

\(\Rightarrow B=115\)

Vậy \(B=115\)