Question

Tính biểu thức B 

Answer 1

\(B=1.2^2+2.3^2+...+99.100^2\\=(2-1).2^2+(3-1).2^2+(100-1).100^2\\=2^3-2^2+3^3-3^2+...+100^3-100^2\\ =(2^3+3^3+...+100^3)-(2^2+3^2+...+100^3)\\ =(1^3+2^3+3^3+...+100^3)-(1^2+2^2+3^2+...+100^3)\\ =\dfrac{[100.(100+1)]^2}{4}-\dfrac{100.(100+1).(2.100+1)}{6} =25164150\)