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\[\large \sin^{-1}\dfrac{1}{\sqrt{2}} + \sin^{-1}\dfrac{\sqrt{2} - 1}{\sqrt{6}} + \sin^{-1}\dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{12}} + \cdots= \ ? \] I am not able to generalise the series..

This is a MCQ type question and options are :

  • 0
  • 1
  • \(\pi / 2\)
  • 2
  • None of these

Note by Akhil Bansal
12 months ago

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\[\sum_{n=1}^{\infty}sin^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}})\] The other side in a right angled Triangle with other sides \(\sqrt{n(n+1)},\sqrt{n}-\sqrt{n-1}\) is \(\sqrt{n^{2}-n}+1\) Therefore \[\sum_{n=1}^{\infty}sin^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}})= \sum_{n=1}^{\infty} tan^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{1+\sqrt{n(n-1)}})\] \[= \sum_{n=1}^{\infty} tan^{-1}(\sqrt{n})-tan^{-1}(\sqrt{n-1})\] \[= tan^{-1}(\infty)-tan^{-1}(0)\] \[=\frac{\pi}{2}\]. Shivam Jadhav · 12 months ago

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@Shivam Jadhav nice one. dont give \(\infty\) in an expression like that. arithmetic for \(\infty\) is not defined. write as \[\lim_{n\rightarrow \infty} (\tan^{-1}(n)-\tan^{-1}(0))\] Aareyan Manzoor · 11 months, 4 weeks ago

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Notist that sin(arcsin(1/√2)+arcsin((√2-1)/√6)= 1/√2(√2-1)/√6+1/√2(√2+1)/√6= 2√2/√12=√(2/3) So sum first two terms is arcsin(√(2/3)) Now sin(arcsin(√(2/3)+arcsin((√3-√2)/√12)=√(2/3)(√6+1)/√12+ +√(1/3)(√3-√2)/√12= 1/√36*(√12+√2+√3-√2)=3√3/6= √(3/4) so using mathematical induction you can prove that sum of first n terms is arccos(√(n/(n+1))= arccos(√(1-1/(n+1)) So when you put lim n->infinify You'll get arcsin(1)=π/2 Nikola Djuric · 12 months ago

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Comment deleted 12 months ago

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@Satyendra Kumar What is the general term to given expression? Akhil Bansal · 12 months ago

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@Akhil Bansal @Akhil Bansal The general term is \(\sin^{-1} \left( \dfrac{\sqrt n -\sqrt{n-1}}{\sqrt{n(n+1)}} \right) \). Sandeep Bhardwaj · 12 months ago

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@Sandeep Bhardwaj Thanks Sir, now I will do the rest. Akhil Bansal · 12 months ago

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