$f(x)=(c+1) x^{2}+\left(1-c^{2}\right) x+2 k$        $\dots(1)$

$f(x+y)=f(x)+f(y)-x y \quad \forall x y \in R$

$\lim \limits_{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}=\lim \limits_{y \rightarrow 0} \frac{f(y)-x y}{y}\Rightarrow f^{\prime}(x)=f^{\prime}(0)-x$

$f(x)=-\frac{1}{2} x^{2}+f^{\prime}(0) \cdot x+\lambda \quad \text { but } f(0)=0 \Rightarrow \lambda=0$

$f(x)=-\frac{1}{2} x^{2}+\left(1-c^{2}\right) \cdot x$        $\dots(2)$

$\therefore \quad$ as $f ^{\prime}(0)=1- c ^{2}$

Comparing equation $(1)$ and $(2)$

We obtain, $c =-\frac{3}{2}$

$\therefore \quad f ( x )=-\frac{1}{2} x ^{2}-\frac{5}{4} x$

Now $\left|2 \sum \limits_{x=1}^{20} f(x)\right|=\sum \limits_{x=1}^{20} x^{2}+\frac{5}{2} \cdot \sum \limits_{x=1}^{20} x$

$=2870+525$

$=3395$