F \(=1.100+2.\left(100-1\right)+3.\left(100-2\right)+...+100\left(100-99\right)\)
\(=1.100+2.100-1.2+3.100-2.3+...+100.100-99.100\)
\(=100\left(1+2+3+...+100\right)-\left(1.2+2.3+3.4+...+99.100\right)\)
\(=100.\frac{101.100}{2}-\frac{99.100.101}{3}\)\(=\)\(505000-333300=171700\)
=> F = 171700
đúng cái nhe