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

$$\mathrm{Find}\mathrm{the}\mathrm{sum}:\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{\tan }^{-1}\left(\frac{2\mathrm{k}}{2+{\mathrm{k}}^2+{\mathrm{k}}^4}\right)$$$$$$$$\mathrm{Answer}:{\tan }^{-1}({\mathrm{n}}^2+\mathrm{n}+1)-\frac{\pi }{4}$$

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

$$T_k={tan}^{-1}\left(\frac{2k}{1+1+k^2+k^4}\right)$$$$now{k}^4+k^2+1$$$$={\left(k^2\right)}^2+2.k^2.1+{\left(1\right)}^2-k^2$$$$={(k^2+1)}^2-{\left(k\right)}^2$$$$=(k^2+1+k)(k^2+1-k)$$$$see$$$$\{(k^2+k+1)-(k^2-k+1)\}$$$$=2k$$$$so{T}_k={tan}^{-1}\left(\frac{2k}{1+k^2+k^4}\right)$$$$={tan}^{-1}\left[\frac{\{(k^2+k+1)-(k^2-k+1)\}}{1+(k^2+k+1)(k^2-k+1)}\right]$$$$={tan}^{-1}(k^2+k+1)-{tan}^{-1}(k^2-k+1)$$$$$$$$T_1={tan}^{-1}\left(3\right)-{tan}^{-1}\left(1\right)$$$$T_2={tan}^{-1}\left(7\right)-{tan}^{-1}\left(3\right)$$$$T_3={tan}^{-1}\left(13\right)-{tan}^{-1}\left(7\right)$$$$....$$$$....$$$$T_n={tan}^{-1}(n^2+n+1)-{tan}^{-1}(n^2-n+1)$$$$addthem$$$$afteradditiononly{tan}^{-1}(n^2+n+1)and{tan}^{-1}\left(1\right)$$$$remainsotherstermscancels$$$$soansweris{S}_n={tan}^{-1}(n^2+n+1)-{tan}^{-1}\left(1\right)$$$$={tan}^{-}(n^2+n+1)-\frac{\pi }{4}$$$$$$$$)$$$$$$

Commented by Tawa1 last updated on 20/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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