int_{0}^{1/3} (sum_{r=0}^{101} {x + frac{r}{3 | vedclass

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(B) We know that $\sum_{r=0}^{n-1} \{x + \frac{r}{n}\} = \{nx\}$.Here,the sum is $\sum_{r=0}^{101} \{x + \frac{r}{3}\}$.Since $\{x + \frac{r}{3}\}$ is

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$17$

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(B) We know that $\sum_{r=0}^{n-1} \{x + \frac{r}{n}\} = \{nx\}$.Here,the sum is $\sum_{r=0}^{101} \{x + \frac{r}{3}\}$.Since $\{x + \frac{r}{3}\}$ is periodic with period $\frac{1}{3}$,we can group the terms in sets of $3$.There are $102$ terms ($r=0$ to $101$),which form $34$ groups of $3$ terms each.Each group is of the form $\{x + \frac{k}{3}\} + \{x + \frac{k+1}{3}\} + \{x + \frac{k+2}{3}\} = \{3x\}$.For $x \in [0, 1/3]$,$\{3x\} = 3x$.Thus,the integral becomes $\int_{0}^{1/3} 34 	imes (3x) dx$.$= 34 	imes 3 \int_{0}^{1/3} x dx = 102 	imes [\frac{x^2}{2}]_{0}^{1/3}$.$= 102 	imes \frac{1}{2} 	imes \frac{1}{9} = \frac{51}{9} = \frac{17}{3}$.Wait,re-evaluating the sum: $\sum_{r=0}^{2} \{x + \frac{r}{3}\} = \{x\} + \{x + \frac{1}{3}\} + \{x + \frac{2}{3}\} = 3x$ for $x \in [0, 1/3]$.There are $34$ such groups. So the sum is $34(3x) = 102x$.$\int_{0}^{1/3} 102x dx = 102 [\frac{x^2}{2}]_{0}^{1/3} = 51 	imes \frac{1}{9} = \frac{51}{9} = 5.66$. Re-checking the sum: $\sum_{r=0}^{101} \{x + \frac{r}{3}\}$. For $x \in [0, 1/3]$,$\{x\}=x, \{x+1/3\}=x+1/3, \{x+2/3\}=x+2/3$. Sum $= 3x+1$.Integral $= \int_{0}^{1/3} 34(3x+1) dx = 34 [\frac{3x^2}{2} + x]_{0}^{1/3} = 34 [\frac{3(1/9)}{2} + 1/3] = 34 [1/6 + 1/3] = 34 [1/2] = 17$.

$\int_{0}^{1/3} (\sum_{r=0}^{101} \{x + \frac{r}{3}\}) dx$ is equal to (where $\{.\}$ represents the fractional part function).

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