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


ANSWER

$$8$$

$$-9$$


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
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