The easiest way to explain this is to say: The first term is #-2# and each number in the sequence after that is the previous one times #-3#. That is called a recursive definition and would look like this:

#f(1)=-2#

#f(n)=F(n-1)*-3#

Once we start with #f(1)# we can just plug it into the second line to get #f(2)=6#, #f(3)=-18#, and so on. But if we want a single equation that will let us know #f(100)# without calculating all 99 previous numbers in the sequence, that formula is pretty much useless.

So let's think about this. Looking at the sequence we see that:

#f(2)=-2*-3= -2*(-3)^1#

#f(3)=-2*-3*-3=-2*(-3)^2#

#f(4)=-2*-3*-3*-3=-2*(-3)^3#

Notice that the exponent that #-3# is getting raised to is 1 less than the sequence number we want. It gets raised to 1 when #n# is 2, and so on. That also works in our favor as #x^0=1# so that #f(1)=-2*(-3)^0#.

That leads us to the final equation:

#f(n)=-2*(-3)^(n-1)#