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If we look at the absolute value of the terms, we see that they follow the pattern that the nth term is 2n.

In order to account for the alternating signs, we are going to need to have the 2n multiplied by (-1)^(n-1), yielding 2n*(-1)^(n-1).

Check the first few terms,

2*1*(-1)^(1-1)=2

2*2*(-1)^(2-1)=-4

2*3*(-1)^(3-1)=6

Looks good.

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