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Question-22940




Question Number 22940 by selestian last updated on 24/Oct/17
Commented by ajfour last updated on 24/Oct/17
T_r =tan^(−1) ((2^(r−1) /(1+2^(2r−1) )))  ⇒  tan T_r =((2^r −2^(r−1) )/(1+2^r .2^(r−1) ))                   =tan (θ_r −θ_(r−1) )      such that   tan θ_r =2^r   So     T_r =θ_r −θ_(r−1)                  =tan^(−1) (2^r )−tan^(−1) (2^(r−1) )           Σ_(r=1) ^n T_r  =tan^(−1) (2^n )−tan^(−1) 1                      =tan^(−1) (2^n )−(π/4) .
$${T}_{{r}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}^{{r}−\mathrm{1}} }{\mathrm{1}+\mathrm{2}^{\mathrm{2}{r}−\mathrm{1}} }\right) \\ $$$$\Rightarrow\:\:\mathrm{tan}\:{T}_{{r}} =\frac{\mathrm{2}^{{r}} −\mathrm{2}^{{r}−\mathrm{1}} }{\mathrm{1}+\mathrm{2}^{{r}} .\mathrm{2}^{{r}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{tan}\:\left(\theta_{{r}} −\theta_{{r}−\mathrm{1}} \right) \\ $$$$\:\:\:\:{such}\:{that}\:\:\:\mathrm{tan}\:\theta_{{r}} =\mathrm{2}^{{r}} \\ $$$${So}\:\:\:\:\:{T}_{{r}} =\theta_{{r}} −\theta_{{r}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}^{{r}} \right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}^{{r}−\mathrm{1}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{T}_{{r}} \:=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}^{{n}} \right)−\mathrm{tan}^{−\mathrm{1}} \mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}^{{n}} \right)−\frac{\pi}{\mathrm{4}}\:. \\ $$

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