For x in mathbb{R},let [x] denote the greates | vedclass

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(C) Let $S = \sum_{k=0}^{99} \left[ -\frac{1}{3} - \frac{k}{100} \right]$.We know that $[x] = -1$ if $-1 \le x < 0$ and $[x] = -2$ if $-2 \le x < -1$.

Vedclass · Accepted Answer

$-133$

Vedclass · Answer

(C) Let $S = \sum_{k=0}^{99} \left[ -\frac{1}{3} - \frac{k}{100} ight]$.We know that $[x] = -1$ if $-1 \le x For the term $\left[ -\frac{1}{3} - \frac{k}{100} ight]$,the value is $-1$ when $-\frac{1}{3} - \frac{k}{100} \ge -1$,which implies $\frac{k}{100} \le 1 - \frac{1}{3} = \frac{2}{3}$,so $k \le \frac{200}{3} \approx 66.66$.Thus,for $k = 0, 1, 2, \dots, 66$ (total $67$ terms),the value is $-1$.For $k = 67, 68, \dots, 99$ (total $99 - 67 + 1 = 33$ terms),the value is $-2$.Sum $= 67 	imes (-1) + 33 	imes (-2) = -67 - 66 = -133$.

For $x \in \mathbb{R}$,let $[x]$ denote the greatest integer $\le x$. Find the sum of the series $\left[ -\frac{1}{3} \right] + \left[ -\frac{1}{3} - \frac{1}{100} \right] + \left[ -\frac{1}{3} - \frac{2}{100} \right] + \dots + \left[ -\frac{1}{3} - \frac{99}{100} \right]$.

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