We're almost done with this stage of the simplification. So over here we have a common 2mb. So let's put a plus 2mb times, once again, x1 plus x2 plus all the way to xn. So this term right over here is the exact same thing as this term over here. And then finally we have a b squared in each of these. And how many of these b squares do we have? Well, we have n of these lines, right?