Question Number 73818 by rajesh4661kumar@gmail.com last updated on 16/Nov/19
Commented by Tinku Tara last updated on 16/Nov/19
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{post}\:\mathrm{a}\:\mathrm{clear}\:\mathrm{image}. \\ $$
Answered by Tanmay chaudhury last updated on 16/Nov/19
![T_r =tan^(−1) ((1/(1+r+r^2 )))=tan^(−1) [(((r+1)−r)/(1+r(r+1)))]=tan^(−1) (r+1)−tan^(−1) (r) S_n =Σ_(r=1) ^n T_r T_1 =tan^(−1) 2−tan^(−1) 1 T_2 =tan^(−1) 3−tan^(−1) 2 ... ... T_n =tan^(−1) (n+1)−tan^(−1) n add them S_n =tan^(−1) (n+1)−tan^(−1) 1 S_∞ =tan^(−1) (∞)−tan^(−1) 1 =(π/2)−(π/4)=(π/4) Answer](https://www.tinkutara.com/question/Q73833.png)
$${T}_{{r}} ={tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{r}+{r}^{\mathrm{2}} }\right)={tan}^{−\mathrm{1}} \left[\frac{\left({r}+\mathrm{1}\right)−{r}}{\mathrm{1}+{r}\left({r}+\mathrm{1}\right)}\right]={tan}^{−\mathrm{1}} \left({r}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left({r}\right) \\ $$$${S}_{{n}} =\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{T}_{{r}} \\ $$$${T}_{\mathrm{1}} ={tan}^{−\mathrm{1}} \mathrm{2}−{tan}^{−\mathrm{1}} \mathrm{1} \\ $$$${T}_{\mathrm{2}} ={tan}^{−\mathrm{1}} \mathrm{3}−{tan}^{−\mathrm{1}} \mathrm{2} \\ $$$$… \\ $$$$… \\ $$$${T}_{{n}} ={tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} {n} \\ $$$${add}\:{them} \\ $$$${S}_{{n}} ={tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \mathrm{1} \\ $$$${S}_{\infty} ={tan}^{−\mathrm{1}} \left(\infty\right)−{tan}^{−\mathrm{1}} \mathrm{1} \\ $$$$=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}}\:{Answer} \\ $$
Commented by mind is power last updated on 17/Nov/19
$$\mathrm{nice}\:\mathrm{worck}\:\mathrm{sir} \\ $$
Commented by Tanmay chaudhury last updated on 17/Nov/19
$${thank}\:{you}\:{sir} \\ $$