Mathematics · 2023
JEE Main · 13 April 2023, Shift 1 · Q7
Among (S1): lim_n → ∞ 1/n^2(2+4+6+… …+2 n)=1 (S2): lim_n → ∞ 1/(n^16)(1^15+2^15+3^15+… …+n^15)=1/16
Among
(S1) : $\displaystyle \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots \ldots+2 n)=1$
(S2) : $\displaystyle \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots+n^{15}\right)=\frac{1}{16}$
Official answer
From NTA’s final answer key for this paper.
(3)
Both (S1) and (S2) are true
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.