Question Number 46032 by Tawa1 last updated on 20/Oct/18
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}:\:\:\:\:\underset{\mathrm{k}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\:\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\mathrm{2k}}{\mathrm{2}\:+\:\mathrm{k}^{\mathrm{2}} \:+\:\mathrm{k}^{\mathrm{4}} }\right) \\ $$$$ \\ $$$$\mathrm{Answer}:\:\:\:\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{n}\:+\:\mathrm{1}\right)\:−\:\frac{\pi}{\mathrm{4}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18
![T_k =tan^(−1) (((2k)/(1+1+k^2 +k^4 ))) now 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) see {(k^2 +k+1)−(k^2 −k+1)} =2k so T_k =tan^(−1) (((2k)/(1+k^2 +k^4 ))) =tan^(−1) [(({(k^2 +k+1)−(k^2 −k+1)})/(1+(k^2 +k+1)(k^2 −k+1)))] =tan^(−1) (k^2 +k+1)−tan^(−1) (k^2 −k+1) T_1 =tan^(−1) (3)−tan^(−1) (1) T_2 =tan^(−1) (7)−tan^(−1) (3) T_3 =tan^(−1) (13)−tan^(−1) (7) .... .... T_n =tan^(−1) (n^2 +n+1)−tan^(−1) (n^2 −n+1) add them after addition only tan^(−1) (n^2 +n+1)and tan^(−1) (1) remains others terms cancels so answer is S_n =tan^(−1) (n^2 +n+1)−tan^(−1) (1) =tan^− (n^2 +n+1)−(π/4) )](https://www.tinkutara.com/question/Q46034.png)
$${T}_{{k}} ={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{k}}{\mathrm{1}+\mathrm{1}+{k}^{\mathrm{2}} +{k}^{\mathrm{4}} }\right) \\ $$$${now}\:{k}^{\mathrm{4}} +{k}^{\mathrm{2}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:=\left({k}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}.{k}^{\mathrm{2}} .\mathrm{1}+\left(\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\left({k}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left({k}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\left({k}^{\mathrm{2}} +\mathrm{1}+{k}\right)\left({k}^{\mathrm{2}} +\mathrm{1}−{k}\right) \\ $$$${see} \\ $$$$\left\{\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)−\left({k}^{\mathrm{2}} −{k}+\mathrm{1}\right)\right\} \\ $$$$=\mathrm{2}{k} \\ $$$${so}\:{T}_{{k}} ={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{k}}{\mathrm{1}+{k}^{\mathrm{2}} +{k}^{\mathrm{4}} }\right) \\ $$$$\:\:\:\:\:\:\:={tan}^{−\mathrm{1}} \left[\frac{\left\{\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)−\left({k}^{\mathrm{2}} −{k}+\mathrm{1}\right)\right\}}{\mathrm{1}+\left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} −{k}+\mathrm{1}\right)}\right] \\ $$$$\:\:\:\:={tan}^{−\mathrm{1}} \left({k}^{\mathrm{2}} +{k}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left({k}^{\mathrm{2}} −{k}+\mathrm{1}\right) \\ $$$$ \\ $$$${T}_{\mathrm{1}} ={tan}^{−\mathrm{1}} \left(\mathrm{3}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$${T}_{\mathrm{2}} ={tan}^{−\mathrm{1}} \left(\mathrm{7}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{3}\right) \\ $$$${T}_{\mathrm{3}} ={tan}^{−\mathrm{1}} \left(\mathrm{13}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{7}\right) \\ $$$$…. \\ $$$$…. \\ $$$${T}_{{n}} ={tan}^{−\mathrm{1}} \left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right) \\ $$$${add}\:{them} \\ $$$${after}\:{addition}\:{only}\:\:{tan}^{−\mathrm{1}} \left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right){and}\:{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$${remains}\:{others}\:{terms}\:{cancels} \\ $$$${so}\:{answer}\:{is}\:{S}_{{n}} ={tan}^{−\mathrm{1}} \left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:={tan}^{−} \left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)−\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$\left.\right) \\ $$$$ \\ $$
Commented by Tawa1 last updated on 20/Oct/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$