Answer:
A. m(-1)^(n-1)
Step-by-step explanation:
The degree of m doesn't change so none of the expressions with n as an exponent of m can possibly be correct. That leaves only choice A.
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You will find that when alternating signs are required, the term (-1)^n will be a factor. An offset of ±1 can be added to the n to make the initial sign be what you want. Here we want +1 for n=1, so an exponent of (n-1) is appropriate.
