Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 49952 by maxmathsup by imad last updated on 12/Dec/18

1) simplify A_n = (1/((2+i(√3))^n )) +(1/((2−i(√3))^n )) 2) smplify B_n =(1/((2+(√3))^n )) +(1/((2−(√3))^n )) n integr natural.

$$\left.\mathrm{1}\right)\:{simplify}\:{A}_{{n}} =\:\frac{\mathrm{1}}{\left(\mathrm{2}+{i}\sqrt{\mathrm{3}}\right)^{{n}} }\:+\frac{\mathrm{1}}{\left(\mathrm{2}−{i}\sqrt{\mathrm{3}}\right)^{{n}} } \\ $$$$\left.\mathrm{2}\right)\:{smplify}\:\:{B}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{n}} }\:+\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{n}} } \\ $$$${n}\:{integr}\:{natural}. \\ $$

Commented by Abdo msup. last updated on 14/Dec/18

1) we have A_n = (((2+i(√3))^n +(2−i(√3))^n )/7^n ) =((2Re( (2+i(√3))^n ))/7^n ) but 2+i(√3)=(√7)((2/(√7)) +((i(√3))/(√7))) =r e^(iθ) ⇒ r=(√7)and cosθ =(2/(√7)) ,sinθ =((√3)/(√7)) ⇒tanθ =((√3)/2) ⇒θ =arctan(((√3)/2)) ⇒(2+i(√3))^n =((√7))^n e^(in arctan(((√3)/2))) ⇒((2((√7))^n cos(narctan(((√3)/2))))/7^n ) = (2/(((√7))^n )) cos(n arctan(((√3)/2))).

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{n}} =\:\frac{\left(\mathrm{2}+{i}\sqrt{\mathrm{3}}\right)^{{n}} \:+\left(\mathrm{2}−{i}\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{7}^{{n}} } \\ $$$$=\frac{\mathrm{2}{Re}\left(\:\left(\mathrm{2}+{i}\sqrt{\mathrm{3}}\right)^{{n}} \right)}{\mathrm{7}^{{n}} }\:\:{but}\:\mathrm{2}+{i}\sqrt{\mathrm{3}}=\sqrt{\mathrm{7}}\left(\frac{\mathrm{2}}{\sqrt{\mathrm{7}}}\:+\frac{{i}\sqrt{\mathrm{3}}}{\sqrt{\mathrm{7}}}\right)\:={r}\:{e}^{{i}\theta} \:\Rightarrow \\ $$$${r}=\sqrt{\mathrm{7}}{and}\:{cos}\theta\:=\frac{\mathrm{2}}{\sqrt{\mathrm{7}}}\:,{sin}\theta\:=\frac{\sqrt{\mathrm{3}}}{\sqrt{\mathrm{7}}}\:\Rightarrow{tan}\theta\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\theta\:={arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\left(\mathrm{2}+{i}\sqrt{\mathrm{3}}\right)^{{n}} \:=\left(\sqrt{\mathrm{7}}\right)^{{n}} \:{e}^{{in}\:{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \:\:\:\Rightarrow\frac{\mathrm{2}\left(\sqrt{\mathrm{7}}\right)^{{n}} {cos}\left({narctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)}{\mathrm{7}^{{n}} } \\ $$$$=\:\frac{\mathrm{2}}{\left(\sqrt{\mathrm{7}}\right)^{{n}} }\:{cos}\left({n}\:{arctan}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right). \\ $$$$ \\ $$

Commented by Abdo msup. last updated on 14/Dec/18

2) we hsve B_n =(((2+(√3))^n +(2−(√3))^n )/((2^2 −((√3))^2 )^n )) = Σ_(k=0) ^n C_n ^k 2^(n−k) ((√3))^k +Σ_(k=0) ^n C_n ^k 2^(n−k) (−(√3))^k =Σ_(k=0) ^n C_n ^k { ((√3))^k +(−(√3))^k )2^(n−k) =Σ_(p=0) ^([(n/2)]) C_n ^(2p) {3^p +3^p ) 2^(n−2p) =2^(n+1) Σ_(p=0) ^([(n/2)]) ((3/4))^p C_n ^(2p)

$$\left.\mathrm{2}\right)\:{we}\:{hsve}\:{B}_{{n}} =\frac{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{n}} \:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{n}} }{\left(\mathrm{2}^{\mathrm{2}} \:−\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \right)^{{n}} } \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\mathrm{2}^{{n}−{k}} \:\left(\sqrt{\mathrm{3}}\right)^{{k}} \:+\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\mathrm{2}^{{n}−{k}} \left(−\sqrt{\mathrm{3}}\right)^{{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left\{\:\left(\sqrt{\mathrm{3}}\right)^{{k}} \:+\left(−\sqrt{\mathrm{3}}\right)^{{k}} \right)\mathrm{2}^{{n}−{k}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}} \:\:\:\:\left\{\mathrm{3}^{{p}} \:+\mathrm{3}^{{p}} \right)\:\mathrm{2}^{{n}−\mathrm{2}{p}} \\ $$$$=\mathrm{2}^{{n}+\mathrm{1}} \:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{p}} {C}_{{n}} ^{\mathrm{2}{p}} \: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com