Question

[KHANH QUYNH MAI PHAM]

Tính F

F=1.100+2.99+3.98+4.97+...+99.2+1.100

Answer 1

[Dương Lam Hàng]

\(F=1.100+2.99+3.98+...+99.2+1.100\)

     \(=1.100+2.\left(100-1\right)+3.\left(100-2\right)+....+100.\left(100-99\right)\)

      \(=1.100+2.100-2.1+3.100-3.2+...+100.100-100.99\)

        \(=100.\left(1+2+3+...+100\right)-\left(1.2+2.3+3.4+....99.100\right)\)

         \(=100.\frac{101.100}{2}-\frac{99.100.101}{3}=505000-333300=171700\)

Đúng thì ủng hộ nha!

Answer 2

[Tuấn Nguyễn]

F = 1.100 + 2.(100-1) + 3. (100-2) +...+100.(100-99)

F = 1.100 + 2.100 - 1.2 + 3.100 - 2.3 + ...+ 100.100 - 99.100

F = 100 . ( 1+2+3+...+100) - (1.2+2.3+3.4+...+99.100)

= 100.(101.100:2) - (99.100.101:3) = 505000 - 333300 = 17100