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Question Number 131155 by EDWIN88 last updated on 02/Feb/21
Given { ((a_(n+2) =a_(n+1) +(1/2)a_n )),((a_1 =3 ; a_2 =2)) :} find a_n .
$${Given}\:\begin{cases}{{a}_{{n}+\mathrm{2}} ={a}_{{n}+\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}{a}_{{n}} }\\{{a}_{\mathrm{1}} =\mathrm{3}\:;\:{a}_{\mathrm{2}} =\mathrm{2}}\end{cases} \\ $$$$\:{find}\:{a}_{{n}} . \\ $$
Answered by john_santu last updated on 02/Feb/21
Characteristic equation λ^2 −λ−(1/2)=0 λ = ((1±(√(1+2)))/2) = ((1±(√3))/2) so a_n =C_1 (((1+(√3))/2))^n +C_2 (((1−(√3))/2))^2 { ((a_1 = 3 = (((1+(√3))/2))C_1 +(((1−(√3))/2))C_2 )),((a_2 = 2= (((2+(√3))/2))C_1 +(((2−(√3))/2))C_2 )) :} { (((1+(√3))C_1 +(1−(√3))C_2 =6)),(((2+(√3))C_1 +(2−(√3))C_2 =4)) :} (((1+(√3) 1−(√3))),((2+(√3) 2−(√3))) ) ((C_1 ),(C_2 ) ) = ((6),(4) ) C_1 = ( determinant (((6 1−(√3))),((4 2−(√3))))/(2(√3)))= ((8−2(√3))/(2(√3))) = ((4−(√3))/( (√3))) C_2 = ( determinant (((1+(√3) 6)),((2+(√3) 4)))/(2(√3))) = ((−8−2(√3))/(2(√3)))=−((4+(√3))/( (√3))) ⇔ a_n =(((4−(√3))/( (√3))))(((1+(√3))/2))^n −(((4+(√3))/( (√3))))(((1−(√3))/2))^n
$${Characteristic}\:{equation} \\ $$$$\lambda^{\mathrm{2}} −\lambda−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\lambda\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{2}}}{\mathrm{2}}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${so}\:{a}_{{n}} ={C}_{\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} +{C}_{\mathrm{2}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\begin{cases}{{a}_{\mathrm{1}} \:=\:\mathrm{3}\:=\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right){C}_{\mathrm{1}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right){C}_{\mathrm{2}} }\\{{a}_{\mathrm{2}} =\:\mathrm{2}=\:\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right){C}_{\mathrm{1}} +\left(\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right){C}_{\mathrm{2}} }\end{cases} \\ $$$$\:\begin{cases}{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){C}_{\mathrm{1}} +\left(\mathrm{1}−\sqrt{\mathrm{3}}\right){C}_{\mathrm{2}} =\mathrm{6}}\\{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){C}_{\mathrm{1}} +\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){C}_{\mathrm{2}} =\mathrm{4}}\end{cases} \\ $$$$\begin{pmatrix}{\mathrm{1}+\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\:\:\mathrm{1}−\sqrt{\mathrm{3}}}\\{\mathrm{2}+\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\:\:\mathrm{2}−\sqrt{\mathrm{3}}}\end{pmatrix}\:\begin{pmatrix}{{C}_{\mathrm{1}} }\\{{C}_{\mathrm{2}} }\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{4}}\end{pmatrix} \\ $$$${C}_{\mathrm{1}} =\:\frac{\begin{vmatrix}{\mathrm{6}\:\:\:\:\mathrm{1}−\sqrt{\mathrm{3}}}\\{\mathrm{4}\:\:\:\:\mathrm{2}−\sqrt{\mathrm{3}}}\end{vmatrix}}{\mathrm{2}\sqrt{\mathrm{3}}}=\:\frac{\mathrm{8}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{4}−\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}} \\ $$$${C}_{\mathrm{2}} =\:\frac{\begin{vmatrix}{\mathrm{1}+\sqrt{\mathrm{3}}\:\:\:\:\:\:\mathrm{6}}\\{\mathrm{2}+\sqrt{\mathrm{3}}\:\:\:\:\:\mathrm{4}}\end{vmatrix}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\:\frac{−\mathrm{8}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{3}}}=−\frac{\mathrm{4}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Leftrightarrow\:{a}_{{n}} =\left(\frac{\mathrm{4}−\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{4}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} \\ $$$$ \\ $$
Answered by mr W last updated on 02/Feb/21
p^2 −p−(1/2)=0 p=((1±(√3))/2) a_n =A(((1+(√3))/2))^(n−1) +B(((1−(√3))/2))^(n−1) a_1 =A+B=3 a_2 =A(((1+(√3))/2))+B(((1−(√3))/2))=2 A(((1+(√3))/2))+(3−A)(((1−(√3))/2))=2 ⇒A=((9+(√3))/6) ⇒B=3−((9+(√3))/6)=((9−(√3))/6) ⇒a_n =(((9+(√3))/6))(((1+(√3))/2))^(n−1) +(((9−(√3))/6))(((1−(√3))/2))^(n−1) or ⇒a_n =(((4(√3)−3)/( 3)))(((1+(√3))/2))^n −(((4(√3)+3)/( 3)))(((1−(√3))/2))^n
$${p}^{\mathrm{2}} −{p}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${p}=\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${a}_{{n}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} +{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \\ $$$${a}_{\mathrm{1}} ={A}+{B}=\mathrm{3} \\ $$$${a}_{\mathrm{2}} ={A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2} \\ $$$${A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\left(\mathrm{3}−{A}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2} \\ $$$$\Rightarrow{A}=\frac{\mathrm{9}+\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\Rightarrow{B}=\mathrm{3}−\frac{\mathrm{9}+\sqrt{\mathrm{3}}}{\mathrm{6}}=\frac{\mathrm{9}−\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\Rightarrow{a}_{{n}} =\left(\frac{\mathrm{9}+\sqrt{\mathrm{3}}}{\mathrm{6}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} +\left(\frac{\mathrm{9}−\sqrt{\mathrm{3}}}{\mathrm{6}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}−\mathrm{1}} \\ $$$${or} \\ $$$$\Rightarrow{a}_{{n}} =\left(\frac{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}}{\:\mathrm{3}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{3}}{\:\mathrm{3}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{n}} \\ $$

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