If $\lbrack x\rbrack$ denotes the greatest integer $\leq x$, then $\left\lbrack \frac{2}{3} \right\rbrack + \left\lbrack \frac{2}{3} + \frac{1}{99} \right\rbrack + \left\lbrack \frac{2}{3} + \frac{2}{99} \right\rbrack + ... + \left\lbrack \frac{2}{3} + \frac{98}{99} \right\rbrack$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
$\left\lbrack \frac{2}{3} + \frac{r}{99} ightbrack = \left\{ \begin{array}{r}
0,\ \ \ \ \ \ \ \ r < 33 \
1,\ \ r \geq 33
\end{array} ight.\$
$	herefore\ \ \ \ \sum_{r = 0}^{98}{\left\lbrack \frac{2}{3} + \frac{r}{99} ightbrack = \sum_{r = 0}^{32}\left\lbrack \frac{2}{3} + \frac{r}{99} ightbrack + \sum_{r = 33}^{98}\left\lbrack \frac{2}{3} + \frac{r}{99} ightbrack}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 0 + 66 = 66$

Q. If $\lbrack x\rbrack$ denotes the greatest integer $\leq x$, then $\left\lbrack \frac{2}{3} \right\rbrack + \left\lbrack \frac{2}{3} + \frac{1}{99} \right\rbrack + \left\lbrack \frac{2}{3} + \frac{2}{99} \right\rbrack + ... + \left\lbrack \frac{2}{3} + \frac{98}{99} \right\rbrack$ is equal to

Explanation

Reviewed by Dr. Abubacker

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