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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
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 = \(505000-333300=171700\)
=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
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
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