Question

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

• A. 2010
• B. 2009
• C. 2011
• D. 2005

Answer 1

The correct option is A 2010

${\sum }_{k=1}^{4020}[\frac{1}{2}+\frac{k-1}{4020}]=[\frac{1}{2}+0]+[\frac{1}{2}+\frac{1}{4020}]+.....[\frac{1}{2}+\frac{2010}{4020}]+.....[\frac{1}{2}+\frac{4019}{4020}]=0+0+0+....+0+1+1+....+1=\underset{\left(2010times\right)}{\overset{0+0+0+....+0+}{\to }}+\underset{\left(2010times\right)}{\overset{1+1+1+....+1}{\to }}=2010$