Let  $f\left(x\right)=2x^2-x-1$ and  $S=\{n\in Z:\left|f\left(n\right)\right|\le 800\}$. Then, the value of  ${\sum }_{n\in S}f\left(n\right)$  is equal to $\lambda$ then the value of  $\left[\frac{\lambda }{40}\right]$of   (where $\left[.\right]$  denotes GIF)  is equals ___________

Given,  $\left|f\left(n\right)\right|\le 800$ and
$\begin{array}{rr} & f\left(n\right)=2n^2-n-1\\ & \Rightarrow -800\le f\left(n\right)\le 800\\ & \Rightarrow f\left(n\right)+800\ge 0\text{ and }f\left(n\right)-800\le 0\\ & \text{ When }f\left(n\right)+800\ge 0\\ & \Rightarrow 2n^2-n-1+800\ge 0\\ & \Rightarrow 2n^2-n+799\ge 0\end{array}$
It is true for all  $n\in Z$  because   $D<0$. When
$\begin{array}{rr} & f\left(n\right)-800\le 0\\ & \Rightarrow 2n^2-n-1-800\le 0\\ & \Rightarrow 2n^2-n-801\le 0\\ & \Rightarrow n(2n-1)\le 801\\ & \Rightarrow n=-19,-18,\dots ,18,19,20\\ & \therefore \underset{n\in S}{ }f\left(n\right)=\underset{n=-19}{\overset{20}{ }}2n^2-n-1\\ & =4[1^2+2^2+\dots .+{19}^2]+2\times {20}^2-20-40\\ & =\frac{4\times 19\times 20\times 39}{6}+800-60=10620\end{array}$