Cho 2017 số a1,a2,..............,a2017 thỏa mãn:
\(\hept{\begin{cases}a_1+4a_2+..........+2017^2a_{2017}=1\\4a_1+9a_2+............+2018^2a_{2017}=12\\9a_1+16a_2+............+2019^2a_{2017=123}\end{cases}}\)
Tính M=\(16a_1+25a_2+...........+2020^2a_{2017}\)