Changing from Positive to Negative I may mess up a little bit...Sorry for that!
When we want a summation to go negative in case of odd numbers and positive otherwise , we use:
$$\sum\limits_{i=1}^{12} \color{red}{{(-1)}^i} 2x^3 $$
But what if I want it to go positive for say $i=1$ to $i=8$ and then turn negative Without using:
$$\sum\limits_{i=1}^{8} 2x^3 \space + \space \sum\limits_{i=9}^{12} -2x^3$$

I mean i want to use the condition in just one summation instead of splitting things up...Then what to do?
$$\sum\limits_{i=1}^{12} {(-1)}^{\color{blue}{?}} 2x^3$$

Thanks a lot!

A cheeky way of doing it: define $f(i) = \begin{cases} 0 & i \leq 8 \\ 1 & \text{otherwise} \end{cases}$, and plug in $? = f(i)$. But that's lame.
One thing you can do that might be what you're looking for but will definitely make whatever you're doing less understandable is to play with the floor function. If you don't know, $\lfloor x \rfloor$, the floor of $x$, is just $x$ rounded down to the greatest integer below it. In this case, $? = \lfloor i/9 \rfloor$ would do what you want it to do, but it feels like cheating.
