Question

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

Answer 1

$ f ( x )=2 x ^2- x -1 $
$ | f ( x )| \leq 800 $
$ 2 n ^2- n -801 \leq 0$
$ n^2-\frac{1}{2} n-\frac{801}{2} \leq 0$
$ \left(n-\frac{1}{4}\right)^2-\frac{801}{2}-\frac{1}{16} \leq 0$
$ \left(n-\frac{1}{4}\right)^2-\frac{6409}{16} \leq 0 $
$ \left(n-\frac{1}{4}-\frac{\sqrt{6409}}{4}\right)\left(n-\frac{1}{4}+\frac{\sqrt{6409}}{16}\right) \leq 0 $
$\frac{1-\sqrt{6409}}{4} \leq n \leq \frac{1+\sqrt{6409}}{4} $
$ n=\{-19,-18-17, \ldots \ldots . .0,1,2, \ldots \ldots, 20\}$
$\displaystyle \sum_{n \in S} f(x)=\sum\left(2 x^2-x-1\right) $
$ =2\left[19^2+18^2+\ldots . .+1^2+1^2+2^2+\ldots .+19^2+20^2\right]$
$ =4\left[1^2+2^2+\ldots .+19^2\right]+2\left[20^2\right]-20-40$
$ =\frac{4 \times 19 \times 20 \times(2 \times 19+1)}{6}+2 \times 400-60$
$ =\frac{4 \times 19 \times 20 \times 39}{6}+800-60-9880+800-60 $
$=10620$