For $a \in R, |a| > 1$, let $\lim\limits_{x \to \infty} \left( \dfrac{1 + \sqrt[3]{2} + \ldots + \sqrt[3]{n}}{n^{7/3}\left( \dfrac{1}{(an+1)^{2}} + \dfrac{1}{(an+2)^{2}} + \ldots + \dfrac{1}{(an+n)^{2}} \right)} \right) = 54$ Then the possible value(s) of $a$ is/are

For a ∈ R, |a| > 1, let limlimits_(x → ∞) ( (1 + √[3]2 + … + √[3]n)/(n^(7/3)( 1/((an+1)²) + 1/((an+2)²) + … + 1/((an+n)²) )) ) = 54 Then the possible value(s) of a is/are $a \in \mathbb{R}, \quad |a| > 1$ $\lim_{x\to\infty}\frac{1+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{n^{7/3}\left(\frac{1}{(an+1)^2}+\ldots+\frac{1}{(an+n)^2}\right)} = 54$ $\text{Find the possible values of } a$ $\sum_{k=1}^{n} k^{1/3} \sim \int_0^n x^{1/3}\,dx = \frac{3}{4}n^{4/3}$ $\sum_{k=1}^{n}\frac{1}{(an+k)^2} = \frac{1}{n^2}\sum_{k=1}^{n}\frac{1}{\left(a+\frac{k}{n}\right)^2}$ $\sim \frac{1}{n}\int_0^1 \frac{dx}{(a+x)^2} = \frac{1}{n}\cdot\frac{1}{a(a+1)}$ $n^{7/3}\sum_{k=1}^{n}\frac{1}{(an+k)^2} \sim n^{7/3}\cdot\frac{1}{n}\cdot\frac{1}{a(a+1)} = \frac{n^{4/3}}{a(a+1)}$ $\text{Ratio} \sim \frac{\frac{3}{4}n^{4/3}}{n^{4/3}/(a(a+1))} = \frac{3}{4}a(a+1)$ $\frac{3}{4}a(a+1) = 54 \Rightarrow a(a+1) = 72$ $a^2 + a - 72 = 0 \Rightarrow (a-8)(a+9) = 0$ $a = 8 \ \text{or}\ a = -9 \quad (\text{both satisfy } |a| > 1)$ $\text{Correct options: } (1)\ -9 \text{ and } (4)\ 8$