Sum of floor function of general term While doing some solution to get a general term for some series I encountered at the end with this.
$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$ 
, where $i> \dfrac{n}{2}$. How can I get the general term from this summation by geting rid of the floor function. 

What's your problem with floor function?
However you can't get rid of it, but this may help:
$$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$$
$$ = \left(\dfrac{n^2+n-2i}{2n} + \dfrac{n^2+n-2i}{2n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$$
$$ = \left(\dfrac{n^2+n-2i}{2n} + \epsilon({n^2+n-2i}{2n})\right) \left(\dfrac{n^2+n-2i}{2n} - \epsilon({n^2+n-2i}{2n}) +1 \right)$$
When the $\epsilon$ function returns the decimal part instead of integer part of given number.
