Question

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

• A. -9
• B. -6
• C. 7
• D. 8

Answer 1

Answer: (A), (D)

$\underset{n\to \infty }{\lim }$$\frac{n^{1/3}\left(\sum _{r=1}^n{\left(\frac{r}{n}\right)}^{1/3}\right)}{n^{7/3}\left(\sum _{r=1}^n\frac{1}{{\left(an+r\right)}^2}\right)}=54\Rightarrow$$\underset{n\to \infty }{\lim }$ $\left(\frac{\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^{1/3}}{\frac{1}{n}\sum _{r=1}^n\frac{1}{{\left(a+r/n\right)}^2}}\right)=54$ $\Rightarrow$ $\frac{\underset{0}{\overset{1}{\int }}{x}^{1/3}dx}{\underset{0}{\overset{1}{\int }}\frac{1}{{\left(a+x\right)}^2}dx}=54$ $\Rightarrow \frac{\frac{3}{4}}{\frac{1}{a\left(a+1\right)}}=54$
$\Rightarrow a\left(a+1\right)=72\Rightarrow a^2+a-72=0\Rightarrow a=-9,8$