Mathematics - Sequence & Series Question with Solution | TestHub

Given,

$f\left(x\right)=2x^2-x-1$ and $\left|f\left(n\right)\right|\le 800$

So, $2n^2-n-801\le 0$

$\Rightarrow n^2-\frac{1}{2}n-\frac{801}{2}\le 0$

$\Rightarrow {\left(n-\frac{1}{4}\right)}^2-\frac{801}{2}-\frac{1}{16}\le 0$

$\Rightarrow {\left(n-\frac{1}{4}\right)}^2-\frac{6409}{16}\le 0$

$\Rightarrow \left(n-\frac{1}{4}-\frac{\sqrt{6409}}{4}\right)\left(n-\frac{1}{4}+\frac{\sqrt{6409}}{16}\right)\le 0$

$\Rightarrow \frac{1-\sqrt{6409}}{4}\le n\le \frac{1+\sqrt{6409}}{4}$

Since given $n$ is integer,

So, $n=\left\{-19,-18-17,\dots \dots ..0,1,2,\dots \dots ,20\right\}$

Now $\sum _{n\in S}f\left(x\right)=\sum \left(2x^2-x-1\right)$

$=\left(2\sum x^2-\sum x-\sum 1\right)$

$=2\left[{19}^2+{18}^2+\dots .+1^2+1^2+2^2+\dots .+{19}^2+{20}^2\right]-20-40$ $\left(\text{as }\sum x=20 \& \sum 1=40\right)$

$=4\left[1^2+2^2+\dots ..+{19}^2\right]+2\left[{20}^2\right]-20-40$

$=\frac{4\times 19\times 20\times \left(2\times 19+1\right)}{6}+2\times 400-60$

$=\frac{4\times 19\times 20\times 39}{6}+800-60-9880+800-60$

$=10620$