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Question: ![](https://cdn.pureessence.tech/canvas_512.png?top_left_x=547&top_left_y=282&width=300&height=300)\...

12n–12\frac{1}{\sqrt{2n–1^{2}}} + 14n–22\frac{1}{\sqrt{4n–2^{2}}} + ...... 16n–32\frac{1}{\sqrt{6n–3^{2}}} + .........+1n\frac{1}{n} is equal to :

A

π4\frac{\pi}{4}

B

π2\frac{\pi}{2}

C

π6\frac{\pi}{6}

D

π3\frac{\pi}{3}

Answer

π2\frac{\pi}{2}

Explanation

Solution

lim⁡n→∞\lim _ { n \rightarrow \infty } 12n−n2\frac { 1 } { \sqrt { 2 n - n ^ { 2 } } } + 12.2n−22\frac { 1 } { \sqrt { 2.2 n - 2 ^ { 2 } } } +

+ ..........

lim⁡n→∞\lim _ { n \rightarrow \infty }

lim⁡n→∞\lim _ { n \rightarrow \infty }

= ∫01dx2x−x2\int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { 2 x - x ^ { 2 } } } (put r/n = x)

= ∫01dx1−(x−1)2\int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { 1 - ( x - 1 ) ^ { 2 } } }

= = π2\frac { \pi } { 2 }