Find-the-sum-k-1-n-tan-1-2k-2-k-2-k-4-Answer-tan-1-n-2-n-1-pi-4-

Question Number 46032 by Tawa1 last updated on 20/Oct/18

Find the sum : \underset{k = 1}{\sum^{n}} \tan^{- 1} (\frac{2 k}{2 + k^{2} + k^{4}})

Answer : \tan^{- 1} (n^{2} + n + 1) - \frac{π}{4}

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18

T_{k} = {t a n}^{- 1} (\frac{2 k}{1 + 1 + k^{2} + k^{4}})

n o w k^{4} + k^{2} + 1

= {(k^{2})}^{2} + 2 . k^{2} . 1 + {(1)}^{2} - k^{2}

= {(k^{2} + 1)}^{2} - {(k)}^{2}

= (k^{2} + 1 + k) (k^{2} + 1 - k)

s e e

{(k^{2} + k + 1) - (k^{2} - k + 1)}

= 2 k

s o T_{k} = {t a n}^{- 1} (\frac{2 k}{1 + k^{2} + k^{4}})

= {t a n}^{- 1} [\frac{{(k^{2} + k + 1) - (k^{2} - k + 1)}}{1 + (k^{2} + k + 1) (k^{2} - k + 1)}]

= {t a n}^{- 1} (k^{2} + k + 1) - {t a n}^{- 1} (k^{2} - k + 1)

T_{1} = {t a n}^{- 1} (3) - {t a n}^{- 1} (1)

T_{2} = {t a n}^{- 1} (7) - {t a n}^{- 1} (3)

T_{3} = {t a n}^{- 1} (13) - {t a n}^{- 1} (7)

\dots .

\dots .

T_{n} = {t a n}^{- 1} (n^{2} + n + 1) - {t a n}^{- 1} (n^{2} - n + 1)

a d d t h e m

a f t e r a d d i t i o n o n l y {t a n}^{- 1} (n^{2} + n + 1) a n d {t a n}^{- 1} (1)

r e m a i n s o t h e r s t e r m s c a n c e l s

s o a n s w e r i s S_{n} = {t a n}^{- 1} (n^{2} + n + 1) - {t a n}^{- 1} (1)

= {t a n}^{-} (n^{2} + n + 1) - \frac{π}{4}

)

Commented by Tawa1 last updated on 20/Oct/18

God bless you sir .

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