Giải thích các bước giải:
$A=2020.\left ( \dfrac{1}{30}+\dfrac{1}{3.50}+\dfrac{1}{5.70}+...+\dfrac{1}{99.1010} \right )\\
=2020.\dfrac{1}{10}\left ( \dfrac{1}{3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101} \right )\\
=202.\dfrac{1}{2}.\left ( \dfrac{2}{1.3} +\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101} \right )\\
=101.\left (\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101} \right )\\
=101.\left ( \dfrac{1}{1}-\dfrac{1}{101} \right )\\
=101.\left ( \dfrac{101}{101}-\dfrac{1}{101} \right )\\
=101.\dfrac{100}{101}\\
=100
=10^2$
Vậy A là một số chính phương
