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 = \(505000-333300=171700\)

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. ( 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

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

=>F=171700