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

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

Professor PC Thomas Classes · Answer

(10620) : Given, $|f(n)| \leq 800$ and $f(n)=2 n^{2}-n-1$\ $\Rightarrow \quad-800 \leq f(n) \leq 800$\ $\Rightarrow f(n)+800 \geq 0$ and $f(n)-800 \leq 0$\ When $f(n)+800 \geq 0$\ $\Rightarrow \quad 2 n^{2}-n-1+800 \geq 0$\ $\Rightarrow \quad 2 n^{2}-n+799 \geq 0$\ It is true for all $n \in Z$ because $D<0$. When $f(n)-800 \leq 0$\ $\Rightarrow \quad 2 n^{2}-n-1-800 \leq 0$\ $\Rightarrow \quad 2 n^{2}-n-801 \leq 0$\ $\Rightarrow \quad n(2 n-1) \leq 801$\ $\Rightarrow \quad n=-19,-18, \ldots, 18,19,20$\ $	herefore \quad \sum_{n \in S} f(n)=\sum_{n=-19}^{20} 2 n^{2}-n-1$\ $=4\left[1^{2}+2^{2}+\ldots .+19^{2}ight]+2 	imes 20^{2}-20-40$\ $=\frac{4 	imes 19 	imes 20 	imes 39}{6}+800-60=10620$\