T=2020/1+2+2020/1+2+3+….+2020/1+2+3+…2019

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T=2020/1+2+2020/1+2+3+….+2020/1+2+3+…2019

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  2. Đáp án:

     $T=2018$

    Giải thích các bước giải:

     $T=\dfrac{2020}{1+2}+\dfrac{2020}{1+2+3}+…+\dfrac{2020}{1+2+3+…+2019}\\
    =2020.\left ( \dfrac{1}{\dfrac{2.3}{2}}+\dfrac{1}{\dfrac{3.4}{2}}+…+\dfrac{1}{\dfrac{2019.2020}{2}} \right )\\
    =2020.\left ( \dfrac{2}{2.3}+\dfrac{2}{3.4}+…+\dfrac{2}{2019.2020} \right )\\
    =4040.\left ( \dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{2019.2020} \right )\\
    =4040.\left ( \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{2019}-\dfrac{1}{2020} \right )\\
    =4040.\left ( \dfrac{1}{2}-\dfrac{1}{2020} \right )\\
    =4040.\left ( \dfrac{1010}{2020}-\dfrac{1}{2020} \right )\\
    =4040.\dfrac{1009}{2020}\\
    =2018$

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