Đáp án:
$A=2020^2.2021^2.1347.13.37.7$
Giải thích các bước giải:
$\text{Áp dụng: }\\ a.1^2+2^2+3^2+..+n^2=\dfrac{n(n+1)(2n+1)}{6}\\ b.1+2+3+..+n=\dfrac{n(n+1)}{2}\\ \rightarrow A=\dfrac{2020(2020+1)}{2}.\dfrac{2020(2020+1)(2.2020+1)}{6}.(65.111+13.13.37)\\ \rightarrow A=\dfrac{2020^2.2021^2.4041}{12}.((5.13).(3.37)+13.13.37)\\ \rightarrow A=\dfrac{2020^2.2021^2.4041}{12}.(13.37)(5.3+13)\\ \rightarrow A=\dfrac{2020^2.2021^2.4041.13.37.7}{3}\\ \rightarrow A=2020^2.2021^2.1347.13.37.7$
