Mathematics · 2019
JEE Main · 10 April 2019, Shift 1 · Q76
lim_n → ∞(((n+1)^1 / 3)/(n^4 / 3)+((n+2)^1 / 3)/(n^4 / 3)+⋯ ⋯+((2 n)^1 / 3)/(n^4 / 3)) is equal to:
$$\lim _{n \rightarrow \infty}\left(\frac{(n+1)^{1 / 3}}{n^{4 / 3}}+\frac{(n+2)^{1 / 3}}{n^{4 / 3}}+\cdots \cdots+\frac{(2 n)^{1 / 3}}{n^{4 / 3}}\right)
$$
is equal to:
Official answer
(1)
$\displaystyle \frac{3}{4}(2)^{4 / 3}-\frac{3}{4}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.