Animated Solution for Mathematics - Quadratic Equations: Let $f(x)=2x^2-x-1$ and $S=\{n\in \mathbb{Z}:|f(n)|\le 800\}$. Then, the value of ${\sum }_{n\in S}f(n)$ is equal to ________.

The problem asks us to consider the function $f(n)=2n^2-n-1$ subject to the constraint $|f(n)|\le 800$, where $n$ is an integer. This inequality is equivalent to the compound inequality $-800\le 2n^2-n-1\le 800$.

First, we examine the lower bound: $2n^2-n-1\ge -800$, which simplifies to $2n^2-n+799\ge 0$.

Calculating the discriminant $D$ of the quadratic $2n^2-n+799$:

Since $D<0$ and the leading coefficient is positive, the expression $2n^2-n+799$ is always positive for all real $n$. Thus, the lower bound is satisfied for all $n\in \mathbb{Z}$.

Next, we address the upper bound: $2n^2-n-1\le 800$, which simplifies to $2n^2-n-801\le 0$. To find the range of $n$, we solve for the roots of $2n^2-n-801=0$ using the quadratic formula:

Given that $\sqrt{6409}\approx 80.05$, the roots are approximately $n\approx 20.26$ and $n\approx -19.76$. Since $n$ must be an integer, the valid range for $n$ is $-19\le n\le 20$.

This range contains exactly $40$ integers.

We are tasked with calculating the sum $S={\sum }_{n=-19}^{20}(2n^2-n-1)$. By the linearity of summation, we express this as:

For the linear term ${\sum }_{n=-19}^{20}n$, we use symmetry. The sum from $-19$ to $19$ is $0$, leaving only the $n=20$ term:

For the constant term, we sum $1$ over $40$ terms:

Now we compute the sum of squares $2{\sum }_{n=-19}^{20}{n}^2$. We split this into $2\left[{\sum }_{n=1}^{19}{n}^2+0^2+{\sum }_{n=1}^{20}{n}^2\right]$. Using the formula ${\sum }_{k=1}^N{k}^2=\frac{N(N+1)(2N+1)}{6}$:

For $N=19$:

For $N=20$:

The total sum of squares is $2\times (2470+2870)=2\times 5340=10680$. Combining all parts, we get:

The final result is 10620.