Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
S = {2^2} + {3^2} + {4^2} + .. + {2019^2}\\
\to S + 1 = 12 + 22 + 32 + 42 + .. + 20192\\
\to S + 1 = 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + ... + 2019(2020 - 1)\\
\to S + 1 = (1.2 + 2.3 + 3.4 + ... + 2019.2020) - (1 + 2 + 3 + ... + 2019))\\
Có :A = 1.2 + 2.3 + 3.4 + .. + 2019.2020\\
\to 3A = 1.2.3 + 2.3.3 + .. + 2019.2020.3\\
\to 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + .. + 2019.2020.(2021 - 2018)\\
\to 3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + .. + 2019.2020.2021\\
- 2018.2019.2020\\
\to 3A = 2019.2020.2021\\
\to A = \frac{{2019.2020.2021}}{3}\\
\to S + 1 = \frac{{2019.2020.2021}}{3} - \frac{{2019.2020}}{2}\\
\to S = \frac{{2019.2020.2021}}{3} - \frac{{2019.2020}}{2} - 1
\end{array}\)
