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Question Number 105673 by I want to learn more last updated on 30/Jul/20

Answered by mathmax by abdo last updated on 30/Jul/20

$$\mathrm{A}=\sum _{\mathrm{k}=1}^{15}\mathrm{f}\left(\mathrm{k}\right)=\sum _{\mathrm{k}=1}^{15}\arctan \left(\frac{1}{{\mathrm{k}}^2+\mathrm{k}+1}\right)=\sum _{\mathrm{k}=1}^{15}\arctan \left(\frac{\mathrm{k}+1-\mathrm{k}}{1+\mathrm{k}(\mathrm{k}+1)}\right)$$$$\mathrm{let}\mathrm{k}={\mathrm{tanu}}_{\mathrm{k}}\Rightarrow \frac{\mathrm{k}+1-\mathrm{k}}{1+\mathrm{k}(\mathrm{k}+1)}=\frac{\tan \left({\mathrm{u}}_{\mathrm{k}+1}\right)-{\mathrm{tanu}}_{\mathrm{k}}}{1+\tan \left({\mathrm{u}}_{\mathrm{k}}\right){\mathrm{tanu}}_{\mathrm{k}+1}}=\tan ({\mathrm{u}}_{\mathrm{k}+1}-{\mathrm{u}}_{\mathrm{k}})$$$$\Rightarrow \arctan \left(\frac{\mathrm{k}+1-\mathrm{k}}{1+\mathrm{k}(\mathrm{k}+1)}\right)={\mathrm{u}}_{\mathrm{k}+1}-{\mathrm{u}}_{\mathrm{k}}\Rightarrow \mathrm{A}=\sum _{\mathrm{k}=1}^{15}({\mathrm{u}}_{\mathrm{k}+1}-{\mathrm{u}}_{\mathrm{k}})$$$$={\mathrm{u}}_2-{\mathrm{u}}_1+{\mathrm{u}}_3-{\mathrm{u}}_2+....+{\mathrm{u}}_{16}-{\mathrm{u}}_{15}={\mathrm{u}}_{16}-{\mathrm{u}}_1=\arctan \left(16\right)-\arctan \left(1\right)$$$$=\arctan \left(16\right)-\arctan \left(1\right)\Rightarrow$$$$\mathrm{tanA}=\tan (\arctan \left(16\right)-\mathrm{ar}4\tan \left(1\right))=\frac{16-1}{1+16\times 1}=\frac{15}{17}$$$$\mathrm{the}\mathrm{answer}\mathrm{is}\mathrm{b}$$

Commented by I want to learn more last updated on 31/Jul/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

Commented by abdomsup last updated on 31/Jul/20

$$youarewelcome$$

Answered by Dwaipayan Shikari last updated on 31/Jul/20

$$f\left(x\right)={tan}^{-1}\frac{1}{x^2+x+1}={tan}^{-1}\frac{x+1-x}{1+x(x+1)}={tan}^{-1}(x+1)-{tan}^{-1}x$$$$A=\underset{n=1}{\sum ^{15}}{tan}^{-1}(x+1)-{tan}^{-1}x={tan}^{-1}2-{tan}^{-1}1+{tan}^{-1}3-{tan}^{-1}2+..$$$$A={tan}^{-1}16-{tan}^{-1}1={tan}^{-1}\frac{15}{17}$$$$tanA=\frac{15}{17}$$

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