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Mathematics Question on Sum of First n Terms of an AP

The sum of the infinite series sin⁡−112+sin⁡−1(2−16)+sin⁡−1(3−212)+...+sin⁡−1(n−n−1n(n+1))\sin^{-1}\frac{1}{\sqrt2}+\sin^{-1}\left(\frac{\sqrt2-1}{\sqrt6}\right)+\sin^{-1}\left(\frac{\sqrt3-\sqrt2}{\sqrt{12}}\right)+...+\sin^{-1}\left(\frac{\sqrt n-\sqrt{n-1}}{\sqrt{n(n+1)}}\right) is

A

π4\frac{\pi}{4}

B

π3\frac{\pi}{3}

C

π2\frac{\pi}{2}

D

π\pi

Answer

π2\frac{\pi}{2}

Explanation

Solution

S=sin−112+sin−12−16+sin−13−212+S = sin^{-1} \frac{1}{\sqrt{2}} + sin^{-1} \frac{\sqrt{2}-1}{\sqrt{6}} + sin^{-1} \frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}} + .....+sin−1(n−n−1n(n+1))..... + sin^{-1} \left(\frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n\left(n+1\right)}}\right) Now, Tn=sin−1(n−n−1n(n+1))T_{n}= sin^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n\left(n+1\right)}}\right) =sin−1[1n1−(1n+1)2−1n+11−(1n)2] = sin^{-1} \left[ \frac{1}{\sqrt{n}} \sqrt{1-\left(\frac{1}{\sqrt{n+1}}\right)^{2}} -\frac{1}{\sqrt{n+1}}\sqrt{1-\left(\frac{1}{\sqrt{n}}\right)^{2}}\right] =sin−11n−sin−11n+1=sin^{-1} \frac{1}{\sqrt{n}} -sin^{-1} \frac{1}{\sqrt{n+1}} [∵sin−1x−sin−1y=sin−1(x1−y2−y1−x2)]\left[ \because sin^{-1} x -sin^{-1}y = sin^{-1}\left(x\sqrt{1-y^{2}} -y \sqrt{1-x^{2}}\right)\right] ∴S=sin−112+(sin−112−sin13)\therefore S = sin^{-1} \frac{1}{\sqrt{2}} + \left(sin ^{-1} \frac{1}{\sqrt{2}} - sin \frac{1}{\sqrt{3}}\right) +(sin−113−sin−114)+.....+∞+ \left(sin^{-1} \frac{1}{\sqrt{3}}-sin^{-1} \frac{1}{\sqrt{4}}\right) + .....+\infty =2sin−112= 2sin^{-1} \frac{1}{\sqrt{2}} =2(π4)= 2\left(\frac{\pi}{4}\right) =π2= \frac{\pi}{2}