Đáp án:
$T=2018$
Giải thích các bước giải:
$T=\dfrac{2020}{1+2}+\dfrac{2020}{1+2+3}+...+\dfrac{2020}{1+2+3+...+2019}\\
=2020.\left ( \dfrac{1}{\dfrac{2.3}{2}}+\dfrac{1}{\dfrac{3.4}{2}}+...+\dfrac{1}{\dfrac{2019.2020}{2}} \right )\\
=2020.\left ( \dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2019.2020} \right )\\
=4040.\left ( \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2019.2020} \right )\\
=4040.\left ( \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020} \right )\\
=4040.\left ( \dfrac{1}{2}-\dfrac{1}{2020} \right )\\
=4040.\left ( \dfrac{1010}{2020}-\dfrac{1}{2020} \right )\\
=4040.\dfrac{1009}{2020}\\
=2018$
