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

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

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

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

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

Đúng thì ủng hộ nha!

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

=1.100 + 2.(100-1) + 3.(100 - 2 ) +..... + 99 .(100 - 98) + 100 .(100 - 99 )

=1.100 + 2. 100 - 1.2 + 3.100 - 3.2 +...+ 99 . 100 - 99 . 98 + 100 .100 - 100.99

= 100 (1 + 2 + 3 +... +100) - (1.2 +3 .2 +...+ 99.98 + 100.99 

=100.100.101 /2 - 99.100.101 / 3 

= 505000 - 333300= 171700 

                                                                Đ/S : 171700

 

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

=>F=171700

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

=(1.100+2.100+3.100+...........+100.100)-(1.2+2.3+3.4+............+99.100)

=99(1+2+3+..........+99)-333300

=99.\(\frac{99.\left(99+1\right)}{2}\)-333300

=490050-33330

=156750

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

=171700

171700 nhé bạn

F = 1.100 + 2.99 + 3.98 + ... + 98.3 + 99.2 + 100.1 = 1.100 + 2.(100-1) + 3.(100-2) + ... + 99.(100-98) + 100.(100-99) = 100.( 1 + 2 + 3 +... +99 + 100) - ( 1.2 + 2.3 + 3.4 + ... + 98.99 + 99.100) \(=100.\dfrac{\left(1+100\right).100}{2}-A\) = 100. 5050 - A Xét dạng tổng quát của A là B = 1.2 + 2.3 + 3.4 + ...+ n.(n+1) Ta có:
3.B = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ n.(n+1).((n+2) - (n-1))
= 1.2.3 + 2.3.4 + 3.4.5 + ... + (n-1).n.(n+1) + n.(n+1).(n+2) - 0.1.2 - 1.2.3 - 2.3.4 - 3.4.5 - ... - (n-1).n.(n+1)
=n.(n+1).(n+2) \(\Rightarrow B=\dfrac{n\left(n+1\right).\left(n+2\right)}{3}\) \(\Rightarrow A=\dfrac{99.100.101}{3}=333300\) Vậy F = \(505000-333300=171700\)

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

= 1 . 100 + 2 . 100 - 1.2 + 3.100 - 2.3 + ... + 100.100 - 99.100

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

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

Vậy F = 171700