JEE MAIN - Mathematics (2022 - 27th July Morning Shift - No. 20)

Let $$f(x)=2 x^{2}-x-1$$ and $$\mathrm{S}=\{n \in \mathbb{Z}:|f(n)| \leq 800\}$$. Then, the value of $$\sum\limits_{n \in S} f(n)$$ is equal to ___________.
Answer
10620

Explanation

$$\because$$ $$\left| {f(n)} \right| \le 800$$

$$ \Rightarrow - 800 \le 2{n^2} - n - 1 \le 800$$

$$ \Rightarrow 2{n^2} - n - 801 \le 0$$

$$\therefore$$ $$n \in \left[ {{{ - \sqrt {6409} + 1} \over 4},{{\sqrt {6409} + 1} \over 4}} \right]$$ and $$n \in z$$

$$\therefore$$ $$n = - 19, - 18, - 17,\,..........,\,19,20.$$

$$\therefore$$ $$\sum {\left( {2{x^2} - x - 1} \right) = 2\sum {{x^2} - \sum {x - \sum 1 } } } $$.

$$ = 2\,.\,2\,.\,\left( {{1^2} + {2^2}\, + \,...\, + \,{{19}^2}} \right) + 2\,.\,{20^2} - 20 - 40$$

$$ = 10620$$

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