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 (integer)
10620
Solution
<p>$\because$ $\left| {f(n)} \right| \le 800$</p>
<p>$\Rightarrow - 800 \le 2{n^2} - n - 1 \le 800$</p>
<p>$\Rightarrow 2{n^2} - n - 801 \le 0$</p>
<p>$\therefore$ $$n \in \left[ {{{ - \sqrt {6409} + 1} \over 4},{{\sqrt {6409} + 1} \over 4}} \right]$$ and $n \in z$</p>
<p>$\therefore$ $n = - 19, - 18, - 17,\,..........,\,19,20.$</p>
<p>$\therefore$ $\sum {\left( {2{x^2} - x - 1} \right) = 2\sum {{x^2} - \sum {x - \sum 1 } } }$.</p>
<p>$$ = 2\,.\,2\,.\,\left( {{1^2} + {2^2}\, + \,...\, + \,{{19}^2}} \right) + 2\,.\,{20^2} - 20 - 40$$</p>
<p>$= 10620$</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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