lim _{n rightarrow infty} frac{1}{n^3+1}+frac | vedclass

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(B) The given expression is $S_n = \sum_{k=1}^{n} \frac{k^2}{n^3+1}$.Since the denominator $n^3+1$ is independent of $k$,we can write $S_n = \frac{1}{

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$\frac{1}{3}$

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(B) The given expression is $S_n = \sum_{k=1}^{n} \frac{k^2}{n^3+1}$.Since the denominator $n^3+1$ is independent of $k$,we can write $S_n = \frac{1}{n^3+1} \sum_{k=1}^{n} k^2$.Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.Thus,$S_n = \frac{n(n+1)(2n+1)}{6(n^3+1)}$.To find the limit as $n \rightarrow \infty$,we evaluate $\lim _{n \rightarrow \infty} \frac{2n^3+3n^2+n}{6n^3+6}$.Dividing the numerator and denominator by $n^3$,we get $\lim _{n \rightarrow \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6 + \frac{6}{n^3}} = \frac{2}{6} = \frac{1}{3}$.

$\lim _{n \rightarrow \infty} \frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\ldots+\frac{n^2}{n^3+1} = $

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