Function Challenge 3!

In the function $f(x)=x^2-x-\frac{1}{2}$ , substitute $x$ with $x+1$ then you have $f(x+1)=x^2+x-\frac{1}{2}$ , notice that $f(x+1)-f(x)=2x$ , so $\begin{aligned}&\displaystyle \sum_{x=1}^{50}{f(2x)}-\displaystyle \sum_{x=1}^{50}{f(2x-1)}\\&=\displaystyle \sum_{x=1}^{50}{[f(2x)-f(2x-1)]}\\&=f(2)-f(1)+f(4)-f(3)+\ldots+f(100)-f(99)\\&=2(1+3+5+7+\ldots+99)\\\\&=2+6+10+14+\ldots+198\\&=\frac{50\times(2+198)}{2} \\&=5000 \end{aligned}$

find the simplify equation of f (2x)- f( 2x-1)....which is 4x-2.... sub x= 1 you get 2....... subx= 2 you get 6.........d=4, a = 2 , n =50...use arimethic progression equation n/2(2a+(n-1)d) you will get the answer