Đáp án:

$S=2^2+3^2+4^2+..+2019^2=\dfrac{2019.2020.2021.2022}{3}-\dfrac{2019.2020}{2}-1$

Giải thích các bước giải:

$S=2^2+3^2+4^2+..+2019^2$

$\rightarrow S+1=1^2+2^2+3^2+4^2+..+2019^2$

$\rightarrow S+1=1.(2-1)+2.(3-1)+3.(4-1)+...+2019(2020-1)$

$\rightarrow S+1=(1.2+2.3+3.4+...+2019.2020)-(1+2+3+...+2019))$

Ta có:

$A=1.2+2.3+3.4+..+2019.2020$

$\rightarrow 3A=1.2.3+2.3.3+..+2019.2020.3$

$\rightarrow 3A=1.2.(3-0)+2.3.(4-1)+..+2019.2020.(2021-2018)$

$\rightarrow 3A=1.2.3-0.1.2+2.3.4-1.2.3+..+2019.2020.2021-2018.2019.2020$

$\rightarrow 3A=2019.2020.2021$

$\rightarrow A=\dfrac{2019.2020.2021}{3}$

$\rightarrow S+1=\dfrac{2019.2020.2021}{3}-\dfrac{2019.2020}{2}$

$\rightarrow S=\dfrac{2019.2020.2021}{3}-\dfrac{2019.2020}{2}-1$