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 = \(505000-333300=171700\)
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 = \(505000-333300=171700\)
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