For a in R, |a| 1,let lim _{n rightarrow in | vedclass

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(A) We are given the limit: $\lim _{n \rightarrow \infty} \frac{\sum_{r=1}^n r^{1/3}}{n^{7/3} \sum_{r=1}^n \frac{1}{(an+r)^2}} = 54$. Dividing the num

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$1, 2$

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(A) We are given the limit: $\lim _{n ightarrow \infty} \frac{\sum_{r=1}^n r^{1/3}}{n^{7/3} \sum_{r=1}^n \frac{1}{(an+r)^2}} = 54$. Dividing the numerator and denominator by $n^{4/3}$,we rewrite the expression as: $\lim _{n ightarrow \infty} \frac{\frac{1}{n} \sum_{r=1}^n (r/n)^{1/3}}{\frac{1}{n} \sum_{r=1}^n \frac{1}{(a+r/n)^2}} = 54$. Using the definition of a definite integral as the limit of a sum,$\lim_{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^n f(r/n) = \int_0^1 f(x) dx$. The numerator becomes $\int_0^1 x^{1/3} dx = [\frac{3}{4} x^{4/3}]_0^1 = \frac{3}{4}$. The denominator becomes $\int_0^1 \frac{1}{(a+x)^2} dx = [-\frac{1}{a+x}]_0^1 = -(\frac{1}{a+1} - \frac{1}{a}) = \frac{1}{a} - \frac{1}{a+1} = \frac{1}{a(a+1)}$. Thus,the equation is $\frac{3/4}{1/(a(a+1))} = 54$,which simplifies to $\frac{3}{4} a(a+1) = 54$. $a(a+1) = 54 	imes \frac{4}{3} = 72$. $a^2 + a - 72 = 0 \Rightarrow (a+9)(a-8) = 0$. So,$a = -9$ or $a = 8$. Both satisfy $|a| > 1$.

For $a \in R, |a| > 1$,let $\lim _{n \rightarrow \infty} \left( \frac{1+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{n^{7/3} \left( \frac{1}{(an+1)^2} + \frac{1}{(an+2)^2} + \ldots + \frac{1}{(an+n)^2} \right)} \right) = 54$. Then the possible value$(s)$ of $a$ is/are:
$(1) 8$ $(2) -9$ $(3) -6$ $(4) 7$

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For $a \in R, |a| > 1$,let $\lim _{n \rightarrow \infty} \left( \frac{1+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{n^{7/3} \left( \frac{1}{(an+1)^2} + \frac{1}{(an+2)^2} + \ldots + \frac{1}{(an+n)^2} \right)} \right) = 54$. Then the possible value$(s)$ of $a$ is/are:$(1) 8$ $(2) -9$ $(3) -6$ $(4) 7$