IlCreatore wrote:

I interpreted the first way.

Thus,

1^^ = 1! - 1 = 0

2^^ = 2! - 4 = -2

3^^ = 3! - 9 = -3

4^^ = 4! - 16 = 18

So, I would say there are two cases in which n^^<0. Now, we have to find which one of the choice has value 2.

Given our prior computations, it is easy to notice that 1^^-2^^ = 0 - (-2) = 2.

Thus, the answer is B!

you are right, i missed it

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