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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

Characteristicequationλ2λ12=0λ=1±1+22=1±32soan=C1(1+32)n+C2(132)2{a1=3=(1+32)C1+(132)C2a2=2=(2+32)C1+(232)C2{(1+3)C1+(13)C2=6(2+3)C1+(23)C2=4(1+3132+323)(C1C2)=(64)C1=|613423|23=82323=433C2=|1+362+34|23=82323=4+33an=(433)(1+32)n(4+33)(132)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

p2p12=0p=1±32an=A(1+32)n1+B(132)n1a1=A+B=3a2=A(1+32)+B(132)=2A(1+32)+(3A)(132)=2A=9+36B=39+36=936an=(9+36)(1+32)n1+(936)(132)n1oran=(4333)(1+32)n(43+33)(132)n

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