Question

Tính:

1+1/2(1+2)+1/3(1+2+3)+...+1/20(1+2+3+...+20)

Answer 1

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

\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{20}\cdot\dfrac{20\cdot21}{2}\)

\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\)

\(=\dfrac{2+3+4+...+21}{2}=\dfrac{\left(21+2\right)+\left(3+20\right)+...+\left(10+13\right)+\left(11+12\right)}{2}\)

\(=\dfrac{23+23+...+23}{2}=\dfrac{23\cdot10}{2}=23\cdot5=115\)