f(x) is polynomial

Put y = 1/x in given functional equation we get

$$f\left( {x + {1 \over x}} \right) = f(x) + f\left( {{1 \over x}} \right) - 1$$

$$ \Rightarrow (c + 1){\left( {x + {1 \over x}} \right)^2} + (1 - {c^2})\left( {x + {1 \over x}} \right) + 2K$$

$$ = (c + 1){x^2} + (1 - {c^2})x + 2K + (c + 1){1 \over {{x^2}}} + (1 - {c^2}){1 \over x} + 2K - 1$$

$$ \Rightarrow 2(c + 1) = 2K - 1$$ ..... (1)

and put $$x = y = 0$$ we get

$$f(0) = 2 + f(0) - 0 \Rightarrow f(0) = 0 \Rightarrow k = 0$$

$$\therefore$$ $$k = 0$$ and $$2c = - 3 \Rightarrow c = - 3/2$$

$$f(x) = - {{{x^2}} \over 2} - {{5x} \over 4} = {1 \over 4}(5x + 2{x^2})$$

$$\left| {2\sum\limits_{i = 1}^{20} {f(i)} } \right| = \left| {{{ - 2} \over 4}\left( {{{5.20.21} \over 2} + {{2.20.21.41} \over 6}} \right)} \right|$$

$$ = \left| {{{ - 1} \over 2}(2730 + 5740)} \right|$$

$$ = \left| { - {{6790} \over 2}} \right| = 3395$$.