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(B,D) The given limit is $L = \lim_{n 	o \infty} \frac{\sum_{r=1}^n r^a}{(n+1)^{a-1} \sum_{r=1}^n (na+r)}$.First,evaluate the denominator sum: $\sum_

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$5$

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(B,D) The given limit is $L = \lim_{n 	o \infty} \frac{\sum_{r=1}^n r^a}{(n+1)^{a-1} \sum_{r=1}^n (na+r)}$.First,evaluate the denominator sum: $\sum_{r=1}^n (na+r) = n^2a + \frac{n(n+1)}{2} = \frac{2n^2a + n^2 + n}{2} = \frac{n^2(2a+1) + n}{2}$.Substituting this into the limit expression:$L = \lim_{n 	o \infty} \frac{\sum_{r=1}^n r^a}{(n+1)^{a-1} \cdot \frac{n^2(2a+1) + n}{2}} = \lim_{n 	o \infty} \frac{2 \sum_{r=1}^n r^a}{(n+1)^{a-1} n^2(2a+1) (1 + \frac{1}{n(2a+1)})}$.Since $(n+1)^{a-1} \approx n^{a-1}$ as $n 	o \infty$,the expression becomes:$L = \lim_{n 	o \infty} \frac{2 \sum_{r=1}^n r^a}{n^{a-1} n^2 (2a+1)} = \lim_{n 	o \infty} \frac{2 \sum_{r=1}^n r^a}{n^{a+1} (2a+1)} = \frac{2}{2a+1} \lim_{n 	o \infty} \frac{1}{n} \sum_{r=1}^n \left(\frac{r}{n}ight)^a$.Using the definition of the definite integral $\int_0^1 x^a dx = \frac{1}{a+1}$:$L = \frac{2}{(2a+1)(a+1)} = \frac{1}{60}$.Thus,$(2a+1)(a+1) = 120 \implies 2a^2 + 3a + 1 = 120 \implies 2a^2 + 3a - 119 = 0$.Solving the quadratic equation using the formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$a = \frac{-3 \pm \sqrt{9 - 4(2)(-119)}}{4} = \frac{-3 \pm \sqrt{9 + 952}}{4} = \frac{-3 \pm \sqrt{961}}{4} = \frac{-3 \pm 31}{4}$.This gives $a = \frac{28}{4} = 7$ or $a = \frac{-34}{4} = \frac{-17}{2}$.Therefore,the values of $a$ are $7$ and $\frac{-17}{2}$.

For $a \in \mathbb{R}$ (the set of all real numbers),$a \neq -1$,if $\lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1}[(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}$,then $a$ is equal to:

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