Let f(x) = 2x^{2} - x - 1 and S = {n in mathb | vedclass

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(A) Given $f(n) = 2n^{2} - n - 1$. We need to find $S = \{n \in \mathbb{Z} : |2n^{2} - n - 1| \leq 800\}$.This implies $-800 \leq 2n^{2} - n - 1 \leq

Vedclass · Accepted Answer

$10620$

Vedclass · Answer

(A) Given $f(n) = 2n^{2} - n - 1$. We need to find $S = \{n \in \mathbb{Z} : |2n^{2} - n - 1| \leq 800\}$.This implies $-800 \leq 2n^{2} - n - 1 \leq 800$.Solving $2n^{2} - n - 1 \leq 800 \implies 2n^{2} - n - 801 \leq 0$. The roots of $2n^{2} - n - 801 = 0$ are $n = \frac{1 \pm \sqrt{1 + 6408}}{4} = \frac{1 \pm \sqrt{6409}}{4}$. Since $\sqrt{6409} \approx 80.05$,the roots are $\approx -19.76$ and $\approx 20.26$.Thus,$n \in \{-19, -18, \dots, 20\}$.Also,$2n^{2} - n - 1 \geq -800 \implies 2n^{2} - n + 799 \geq 0$. The discriminant $D = 1 - 4(2)(799) = 1 - 6392 Thus,$S = \{-19, -18, \dots, 20\}$.We need to calculate $\sum_{n=-19}^{20} (2n^{2} - n - 1) = 2 \sum_{n=-19}^{20} n^{2} - \sum_{n=-19}^{20} n - \sum_{n=-19}^{20} 1$.$\sum_{n=-19}^{20} n^{2} = (19^{2} + 18^{2} + \dots + 1^{2}) + 0^{2} + (1^{2} + 2^{2} + \dots + 20^{2}) = 2 \sum_{k=1}^{19} k^{2} + 20^{2} = 2 \frac{19(20)(39)}{6} + 400 = 2(4940) + 400 = 10280$.$\sum_{n=-19}^{20} n = (-19 - 18 - \dots - 1) + 0 + (1 + 2 + \dots + 19) + 20 = 20$.$\sum_{n=-19}^{20} 1 = 20 - (-19) + 1 = 40$.Sum $= 2(10280) - 20 - 40 = 20560 - 60 = 10500$. Wait,re-evaluating: $\sum_{n=-19}^{20} (2n^2 - n - 1) = 2(2 \sum_{k=1}^{19} k^2 + 20^2) - 20 - 40 = 2(2470 	imes 2 + 400) - 60 = 2(5340) - 60 = 10680 - 60 = 10620$.

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:

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