MCQ
If $[x]$ denotes the greatest integer $\leq x$, then $\left[\frac{2}{3}\right]+\left[\frac{2}{3}+\frac{1}{99}\right]+\left[\frac{2}{3}+\frac{2}{99}\right]+\ldots .+\left[\frac{2}{3}+\frac{98}{99}\right]=$
  • A
    99
  • B
    98
  • 66
  • D
    65

Answer

Correct option: C.
66
(C)
Given expression
$=\sum_{i=0}^{98}\left[\frac{2}{3}+\frac{i}{99}\right]$
$=\sum_{i=0}^{32}\left[\frac{2}{3}+\frac{i}{99}\right]+\sum_{i=33}^{98}\left[\frac{2}{3}+\frac{i}{99}\right]$
$=0+\sum_{i=33}^{98}\left[\frac{2}{3}+\frac{i}{99}\right]$
$\ldots\left[\because \frac{2}{3} \leq \frac{2}{3}+\frac{ i }{99}<1\right.$ for $\left.i =0,1,2, \ldots, 32\right]$
$=66$
$\ldots\left[\begin{array}{c}\because \text { each term in the summation is one or more } \\ \text { but less than } 2 \text { when } i=33,34,35, \ldots, 98\end{array}\right]$

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