Question

If $\left[x\right]$ represents the greatest integer less than or equal to $x$, then $\sum _{k=1}^{4020}[\frac{1}{2}+\frac{k-1}{4020}]$ is equal to

Answer 1

For $k=1,2,3\cdots \cdots$ upto $2010$ , the value $[\frac{1}{2}+\frac{k-1}{4020}]$ is equal to zero because $\frac{k-1}{4020}<\frac{1}{2}$

For $k=2011,2012,\cdots 4020,$ the value of $[\frac{1}{2}+\frac{k-1}{4020}]=1$
$\therefore$ The required sum value is $2010$