Mathematics · 2023
JEE Main · 25 January 2023, Shift 1 · Q65
The value of lim_n → ∞ (1+2-3+4+5-6+… …+(3 n-2)+(3 n-1)-3 n)/(√(2 n^4+4 n+3)-√(n^4+5 n+4)) is:
The value of $\displaystyle \lim _{n \rightarrow \infty} \frac{1+2-3+4+5-6+\ldots \ldots+(3 n-2)+(3 n-1)-3 n}{\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}}$ is :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \frac{3}{2}(\sqrt{2}+1)$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.