Let $$f(x) = 2x^2 - x - 1$$ and $$S = \{n \in \mathbb{Z} : |f(n)| \leq 800\}$$. Then, the value of $$\sum_{n \in S} f(n)$$ is equal to

We need all integers $$n$$ satisfying $$|2n^2-n-1|\le 800$$

Since

$$2n^2-n-1=(2n+1)(n-1)$$ and the parabola opens upward, solve $$-800\le 2n^2-n-1\le 800$$

First,

$$2n^2-n-1\le 800$$

$$2n^2-n-801\le 0$$

The roots are $$n=\frac{1\pm\sqrt{6409}}{4}=\frac{1\pm 80.056\ldots}{4}$$

Hence

$$-19.76\ldots \le n\le 20.26\ldots$$

So the integer values are $$-19\le n\le 20$$

Now check the other inequality:

$$2n^2-n-1\ge -800$$

$$2n^2-n+799\ge 0$$

Its discriminant is $$1-4(2)(799)=1-6392<0$$ and the quadratic opens upward, so it is always positive.

Therefore

$$S=\{-19,-18,\ldots,20\}$$ and $$\sum_{n\in S}f(n)=\sum_{n=-19}^{20}(2n^2-n-1)$$

There are $$40$$ terms.

Also,

$$\sum_{n=-19}^{20}n=\frac{40(-19+20)}{2}=20$$

Further,

$$\sum_{n=-19}^{20}n^2=\sum_{n=1}^{19}n^2+\sum_{n=1}^{20}n^2$$

$$=\frac{19\cdot20\cdot39}{6}+\frac{20\cdot21\cdot41}{6}$$

$$=2470+2870$$

$$=5340$$

Hence

$$\sum_{n\in S}f(n)=2(5340)-20-40$$

$$=10680-60$$

$$=10620$$

Final Answer : $$10620$$