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Given-a-n-2-a-n-1-1-2-a-n-a-1-3-a-2-2-find-a-n-

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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{an+2=an+1+12ana1=3;a2=2findan.
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
Characteristicequationλ2−λ−12=0λ=1±1+22=1±32soan=C1(1+32)n+C2(1−32)2{a1=3=(1+32)C1+(1−32)C2a2=2=(2+32)C1+(2−32)C2{(1+3)C1+(1−3)C2=6(2+3)C1+(2−3)C2=4(1+31−32+32−3)(C1C2)=(64)C1=|61−342−3|23=8−2323=4−33C2=|1+362+34|23=−8−2323=−4+33⇔an=(4−33)(1+32)n−(4+33)(1−32)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
p2−p−12=0p=1±32an=A(1+32)n−1+B(1−32)n−1a1=A+B=3a2=A(1+32)+B(1−32)=2A(1+32)+(3−A)(1−32)=2⇒A=9+36⇒B=3−9+36=9−36⇒an=(9+36)(1+32)n−1+(9−36)(1−32)n−1or⇒an=(43−33)(1+32)n−(43+33)(1−32)n

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