The limn→∞ Sn if Sn=12n+14n2−1+14n2−4+…+13n2+2n−1 is

limn→∞ Sn=limn→∞ 1n14−0+14−1/n2+14−4/n2+…+4−n−1n2=limn→∞ 1n∑r=0n−1 14−(r/n)2=∫01 dx4−x2=sin−1⁡ x201=π6.

The limn→∞ Sn if Sn=12n+14n2−1+14n2−4+…+13n2+2n−1 is

limn→∞ Sn=limn→∞ 1n14−0+14−1/n2+14−4/n2+…+4−n−1n2=limn→∞ 1n∑r=0n−1 14−(r/n)2=∫01 dx4−x2=sin−1⁡ x201=π6.

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The $lim_{n \to \infty} S_{n}$ if $S_{n} = \frac{1}{2 n} + \frac{1}{\sqrt{4 n^{2} - 1}} + \frac{1}{\sqrt{4 n^{2} - 4}} + \dots + \frac{1}{\sqrt{3 n^{2} + 2 n - 1}}$ is

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$π / 2$

$π / 6$

answer is D.

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Detailed Solution

$lim_{n \to \infty} S_{n}$

$\begin{array}{l} = lim_{n \to \infty} \frac{1}{n} [\frac{1}{\sqrt{4 - 0}} + \frac{1}{\sqrt{4 - 1 / n^{2}}} + \frac{1}{\sqrt{4 - 4 / n^{2}}} + \dots + \sqrt{4 - {(\frac{n - 1}{n})}^{2}}] \\ = lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n - 1} \frac{1}{\sqrt{4 - (r / n)^{2}}} = \int_{0}^{1} \frac{d x}{\sqrt{4 - x^{2}}} = {\sin^{- 1} \frac{x}{2}|}_{0}^{1} = \frac{π}{6} . \end{array}$