Question

If $\left\{x\right\}$ and $\left[x\right]$ represent fractional and integral parts of $x$ respectively, then the value of $\left[x\right]+\sum _{r=1}^{2000}\frac{\{x+2\}}{2000}$ is $\frac{a}{5}x$. Find $a$

Answer 1

In $\left[x\right]+\sum _{r=1}^{2000}\frac{\{x+2\}}{2000}$, we know that $\{x+r\}=\left\{x\right\}$
$\left[x\right]+\sum _{r=1}^{2000}\frac{\left\{x\right\}}{2000}$
$\Rightarrow \left[x\right]+[\frac{\left\{x\right\}}{2000}+\frac{\left\{x\right\}}{2000}+....+\text{upto 2000 times}]$
$\Rightarrow \left[x\right]+\frac{2000\left\{x\right\}}{2000}\Rightarrow \left[x\right]+\left\{x\right\}\Rightarrow x$
Thus, $\left[x\right]+\sum _{r=1}^{2000}\frac{\{x+2\}}{2000}=x$
$\Rightarrow a=5$