\( \lim _{n \rightarrow \infty}\left(\frac{1}{n^{3}+1}+\frac{4}{n^{3}+1}+\frac{9}{n^{3}+1}+\ldots .+\frac{n^{2}}{n^{3}+1}\right) \) is equal to (A) 1 (B) \( 2 / 3 \) (C) \( 1 / 3 \) (D) 0

\( \lim _{n \rightarrow \infty}\left(\frac{1}{n^{3}+1}+\frac{4}{n^{3}+1}+\frac{9}{n^{3}+1}+\ldots .+\frac{n^{2}}{n^{3}+1}\right) \) is equal to (A) 1 (B) \( 2 / 3 \) (C) \( 1 / 3 \) (D) 0

Monjur Alahi

5 views

2025-01-04 04:42

2 Answers

Salim Ahmed Kawsar

1/1+n^3 +4/n^3+1

put n equal to 1/0 then 1/1+{1/0}^3

0/1

then answer is D

5 views Answered 2023-02-05 15:29

Salim Ahmed Kawsar

n(n+1)(2n+1)/6(n³+1)

(2n²+ n)/(6n²-6n+6)

(2+1/n)/(6-6/n+6/n²)

n=infinity

==2/6=1/3

5 views Answered 2023-02-05 15:29

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