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Question: If S<sub>n</sub> =\(\left\lbrack \frac{1}{2n} + \frac{1}{\sqrt{4n^{2} - 1}} + \frac{1}{\sqrt{4n^{2} ...

If Sn =[12n+14n2−1+14n2−4+....+13n2+2n−1]\left\lbrack \frac{1}{2n} + \frac{1}{\sqrt{4n^{2} - 1}} + \frac{1}{\sqrt{4n^{2} - 4}} + .... + \frac{1}{\sqrt{3n^{2} + 2n - 1}} \right\rbrack

then lim⁡n→∞\lim_{n \rightarrow \infty}Sn is equal to -

A

π4\frac{\pi}{4}

B

π6\frac{\pi}{6}

C

π3\frac{\pi}{3}

D

π2\frac{\pi}{2}

Answer

π6\frac{\pi}{6}

Explanation

Solution

lim⁡n→∞\lim_{n \rightarrow \infty} Sn

= lim⁡n→∞\lim _ { n \rightarrow \infty } [14n2−0+14n2−12+14n2−22+...+14n2−(n−1)2]\left\lbrack \frac{1}{\sqrt{4n^{2} - 0}} + \frac{1}{\sqrt{4n^{2} - 1^{2}}} + \frac{1}{\sqrt{4n^{2} - 2^{2}}} + ... + \frac{1}{\sqrt{4n^{2} - (n - 1)^{2}}} \right\rbrack

= lim⁡n→∞\lim_{n \rightarrow \infty}

= 12\frac{1}{2} lim⁡n→∞\lim _ { n \rightarrow \infty } ·

=12\frac { 1 } { 2 } ∫0111−(x2)2\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 - \left( \frac { x } { 2 } \right) ^ { 2 } } } dx = 12\frac{1}{2}. [2sin⁡−1x2]01\left[ 2 \sin ^ { - 1 } \frac { x } { 2 } \right] _ { 0 } ^ { 1 } = π6\frac { \pi } { 6 } .

Hence (2) is the correct answer.