Let f(x)=2x^2-x-1 and S=nZ: |f(n)|800.

Question

Let $f(x)=2x^2-x-1$ and $S=\{n\in\mathbb{Z}: |f(n)|\le800\}$. Then the value of $\displaystyle\sum_{n\in S}f(n)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 10620

Accepted Answer

$|2n^2-n-1|\le800 \Rightarrow 2n^2-n-801\le0 \Rightarrow n\in\left[\dfrac{1-\sqrt{6409}}{4},\dfrac{1+\sqrt{6409}}{4}\right]$, giving integers $n=-19,\ldots,20$.

$$\sum_{n=-19}^{20}(2n^2-n-1) = 2\left[2\sum_{k=1}^{19}k^2+20^2\right] - 20 - 40 = 10620.$$

Solutions · 1