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