Finding the $n$th term for the sequence $1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, \dots$ I have tried using a negative exponent.  I need one statement not two, the pattern is 
$$1, \; \frac{1}{2}, \; 3, \; \frac{1}{4}, \; 5, \; \frac{1}{6}, \dots$$
Hint: You can express it by $n^{a_n}$, where the $a_n$ give the appropriate power needed. For $n$ even, you want $a_n=1$ and for $n$ odd, you want $n=-1$ (recall $x^{-y}=1/x^y$).  

So what formula  gives $1,-1,1,-1,\ldots$?
The answer is simple!
$(2n+1),\frac{n}{2}$ where $n$EZ.