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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

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detailed solution

Correct option is D

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

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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