Mathematics

# For $a\in R,|a|>1$, let $\displaystyle \lim_{n\rightarrow \infty}\dfrac{1+\sqrt[3]{2}+........+\sqrt[3]{n}}{n^{7/3}\left(\dfrac{1}{(na+1)^2}+\dfrac{1}{(na+2)^2}+.......+\dfrac{1}{(na+n)^2}\right)}=54$. Then possible value (s) a is/are

$8$

$-9$

##### SOLUTION
$\displaystyle \lim_{n\rightarrow \infty}\dfrac{\dfrac{1}{n}\sum_{r-1}^n\left(\dfrac{r}{n}\right)^{1/3}}{n^{7/3}\left(\dfrac{n^2}{(na+1)^2}+\dfrac{n^2}{(na+2)^2}+.......+\dfrac{1}{(na+1)^2}\right)}=54$
$\dfrac{\int_0^1 x^{1/3}dx}{\int_0^1\dfrac{dx}{(a+x)^2}}=\dfrac{\left[\dfrac{3}{4}X^{4/3}\right]_0^1}{\left[-\dfrac{1}{a+x}\right]_0^{1}}=\dfrac{\dfrac{3}{4}}{\dfrac{1}{a}-\dfrac{1}{a+1}}=54$
$\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{3}{4\times 54}$
$\Rightarrow \dfrac{1}{a(a+1)}=\dfrac{1}{72}$
$a=8$ or $a=-9$

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Multiple Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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Q2 Subjective Medium
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Q4 Subjective Medium
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