Question

For $a\in R,\left|a\right|>1$, let $\underset{n\to \infty }{\lim }\left(\frac{1+\sqrt[3]{2}+\sqrt[3]{3}+....+\sqrt[3]{n}}{n^{{\displaystyle \frac{7}{3}}}(\frac{1}{{(an+1)}^2}+\frac{1}{{(an+2)}^2}+.....+\frac{1}{{(an+n)}^2})}\right)=54$ then the possible value a is/are.

• A. $8$
• B. $-9$
• C. $-6$
• D. $7$

Answer 1

Answer: (A), (B)

$-9$

Explanation for the correct options:

Finding the possible value of $a$:

$$\begin{gathered} \underset{n\to \infty }{\lim }\left(\frac{1+\sqrt[3]{2}+\sqrt[3]{3}+....+\sqrt[3]{n}}{n^{{\displaystyle \frac{7}{3}}}(\frac{1}{{(an+1)}^2}+\frac{1}{{(an+2)}^2}+.....+\frac{1}{{(an+n)}^2})}\right)=54 \\ \underset{n\to \infty }{\lim }\frac{{\displaystyle \frac{1}{n}}{\displaystyle \sum _{r=1}^n}{\left({\displaystyle \frac{r}{n}}\right)}^{{\displaystyle \frac{1}{3}}}}{{\displaystyle \frac{1}{n}}[\frac{n^2}{{(an+1)}^2}+\frac{n^2}{{(an+2)}^2}+.....+\frac{n^2}{{(an+n)}^2}]}=54 \\ \Rightarrow \frac{{\int }_0^1{x}^{{\displaystyle \frac{1}{3}}}dx}{{\int }_0^1{\displaystyle \frac{dx}{{(a+x)}^2}}}=54\mathbf{}\left[\because \frac{r}{n}\to x,\frac{1}{n}\to \to \mathrm{dx}\right] \\ \Rightarrow \frac{{{\left[{\displaystyle \frac{3}{4}}{x}^{{\displaystyle \frac{4}{3}}}\right]}_0}^1}{{{\left[{\displaystyle \frac{-1}{a+x}}\right]}_0}^1}=\frac{3/4}{{\displaystyle \frac{1}{a}-\frac{1}{a+1}}}=54 \\ \Rightarrow \frac{(a+1)-a}{a(a+1)}=\frac{3}{4}\times \frac{1}{54} \\ \Rightarrow \frac{1}{a(a+1)}=\frac{1}{72} \\ \Rightarrow a^2+a-72=0 \\ \Rightarrow a=8or-9 \end{gathered}$$

Hence, correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathbf{\&}\mathbf{}\mathbf{\left(}\mathit{B}\mathbf{\right)}$.