QJEEM22Madf51: JEE Main 2022 Mathematics - Class 11 JEEM22Madf51 Solution: Let and . Then, the value of is equal to ___________.

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

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

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

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

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

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

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

$= 10620$