For x in R,let [x] denote the greatest intege | vedclass

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(C) Let the series be $S = \sum_{k=0}^{99} \left[ -\frac{1}{3} - \frac{k}{100} \right]$.For $0 \le k \le 66$,we have $-\frac{1}{3} - \frac{k}{100} \ge

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$-133$

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(C) Let the series be $S = \sum_{k=0}^{99} \left[ -\frac{1}{3} - \frac{k}{100} ight]$.For $0 \le k \le 66$,we have $-\frac{1}{3} - \frac{k}{100} \ge -\frac{1}{3} - \frac{66}{100} = -\frac{1}{3} - 0.66 = -0.9966$. Since $-1 Sum for these terms $= 67 	imes (-1) = -67$.For $67 \le k \le 99$,we have $-\frac{1}{3} - \frac{k}{100} \le -\frac{1}{3} - \frac{67}{100} = -0.3333 - 0.67 = -1.0033$. Also,$-\frac{1}{3} - \frac{99}{100} = -0.3333 - 0.99 = -1.3233$. Since $-2 Sum for these terms $= 33 	imes (-2) = -66$.Total sum $S = -67 + (-66) = -133$.

For $x \in R$,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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