Question

If $[.]$ represents the greatest integer function, then evaluate the following sum :
$\left[\frac{1}{1}\right]+\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{3}{3}\right]+\left[\frac{1}{4}\right]+\left[\frac{2}{4}\right]+\left[\frac{3}{4}\right]+\left[\frac{4}{4}\right]+\left[\frac{1}{5}\right]+\cdots$
upto the ${2013}^{\text{th}}$ term.
(correct answer + 2, wrong answer 0)

Answer 1

For $1\le r<k,\left[\frac{r}{k}\right]=0$
The total number of terms upto $\left[\frac{N}{N}\right]$ is $\frac{1}{2}N(N+1)$, and we have the inequality $\frac{62\times 63}{2}=1953<2013<2016=\frac{63\times 64}{2}$
So, the ${2013}^{\text{th}}$ term is $\left[\frac{62}{63}\right]$
$\therefore$ Sum $=1+1+1+\cdots 62\text{times}=62$