Question

If $\left\{x\right\}$ and $\left[x\right]$ represent the fractional and the integral part of $x$ respectively, then $\frac{2019}{2020}\left[x\right]+\frac{x}{2020}+\sum _{r=1}^{2019}\frac{\{x+r\}}{2020}$ is equal to

• A. $x$
• B. $2021x$
• C. $1010x$
• D. $\frac{1}{2021}x$

Answer 1

The correct option is A $x$

We know $x=\left[x\right]+\left\{x\right\}$

Now, $\frac{2019}{2020}\left[x\right]+\frac{x}{2020}+\sum _{r=1}^{2019}\frac{\{x+r\}}{2020}$
$=\frac{2019}{2020}\left[x\right]+\frac{\left[x\right]+\left\{x\right\}}{2020}+\sum _{r=1}^{2019}\frac{\left\{x\right\}}{2020}[\because \{x+r\}=\{x\}\forall r\in Z]$
$=\frac{2019}{2020}\left[x\right]+\frac{\left[x\right]+\left\{x\right\}}{2020}+\frac{\left\{x\right\}}{2020}\sum _{r=1}^{2019}1$
$=\left(\frac{2019}{2020}\right[x]+\frac{1}{2020}[x\left]\right)+\left(\frac{1}{2020}\right\{x\}+\frac{2019}{2020}\{x\left\}\right)$
$=\frac{2020}{2020}\left[x\right]+\frac{2020}{2020}\left\{x\right\}=\left[x\right]+\left\{x\right\}=x$