Question

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

Answer 1

Answer: 10620

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