A number theory problem by Mardokay Mosazghi

The series is actually (2-1)+(3-2)+(4-3)+(5-4)+........2015 terms .Clearly the answer is 2015

sum of even numbers- sum of odd numbers. So is sum of even is = $n^{2}+n$

Sum odd numbers= $n^{2}.$

So plug 2015 for n,

= $(2015^{2}+2015)-(2015^{2})$ = $2015$ $\Rightarrow \boxed{\mathrm\ 2015}$ .

Sum=(2-1)+(4-3)+(6-5)+++++++++

Every pair will leave one, there are 2015 pairs

so ans is 2015!