If Sn-r-1nsin -1-r-r-1-rr-1- then

Sin ^ 1(√(r) √(r 1)√(r(r+1)))=tan ^ 1(√(r) √(r 1)1+√(r)√((r 1))) =tan ^ 1√(r) tan ^ 1(√(r 1)) ⇒ S_n= ∑_r=1^nsin ^ 1(√(r) √(r 1)√(r)(r+1)) ⇒ S_n=∑_r=1^n(tan ^ 1√ ...

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Standard XII

Mathematics

Integration by Parts

If Sn=∑r=1nsi...

Question

If $S_{n} = n \sum r = 1 {sin}^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{\sqrt{r (r + 1)}})$ , then

S_{\infty} = \frac{π}{2}

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S_{3} = \frac{π}{6}

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S_{\infty} = \frac{π}{4}

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S_{3} = \frac{π}{3}

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Solution

The correct option is D $S_{3} = \frac{π}{3}$
${sin}^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{\sqrt{r (r + 1)}}) = {tan}^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{1 + \sqrt{r} \sqrt{(r - 1)}})$
$= {tan}^{- 1} \sqrt{r} - {tan}^{- 1} (\sqrt{r - 1})$
$\Rightarrow S_{n} = n \sum r = 1 {sin}^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{\sqrt{r} (r + 1)})$
$\Rightarrow S_{n} = n \sum r = 1 ({tan}^{- 1} \sqrt{r} - {tan}^{- 1} \sqrt{r - 1})$
$∴ S_{n} = {tan}^{- 1} \sqrt{n}$

Hence $S_{3} = {tan}^{- 1} \sqrt{3} = \frac{π}{3}$
and $S_{\infty} = {tan}^{- 1} \sqrt{\infty} = \frac{π}{2}$

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Similar questions

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Integration by Parts

Standard XII Mathematics

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If Sn=n∑r=1sin−1(√r−√r−1√r(r+1)), then

The correct option is D S3=π3sin−1(√r−√r−1√r(r+1))=tan−1(√r−√r−11+√r√(r−1)) =tan−1√r−tan−1(√r−1) ⇒Sn=n∑r=1sin−1(√r−√r−1√r(r+1)) ⇒Sn=n∑r=1(tan−1√r−tan−1√r−1) ∴Sn=tan−1√n Hence S3=tan−1√3=π3 and S∞=tan−1√∞=π2

If $S_{n} = n \sum r = 1 {sin}^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{\sqrt{r (r + 1)}})$ , then