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Question Number 207060 by efronzo1 last updated on 06/May/24

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Answered by mr W last updated on 05/May/24

Method II a_(n+2) −a_(n+1) =−(1/2)(a_(n+1) −a_n ) let b_n =a_(n+1) −a_n b_(n+1) =−(1/2)b_n b_n =(−(1/2))b_(n−1) =(−(1/2))^2 b_(n−2) =...=(−(1/2))^(n−4) b_4 a_(n+1) −a_n =(−(1/2))^(n−4) (a_5 −a_4 )=(−(1/2))^(n−4) a_(n+1) −a_n =(−(1/2))^(n−4) =16(−(1/2))^n a_n −a_(n−1) =16(−(1/2))^(n−1) a_(n−1) −a_(n−2) =16(−(1/2))^(n−2) ... a_5 −a_4 =16(−(1/2))^4 a_n −a_4 =16(−(1/2))^4 [1+(−(1/2))+(−(1/2))^2 +...+(−(1/2))^(n−5) ] a_n −a_4 =((1−(−(1/2))^(n−4) )/(1+(1/2))) a_n =(2/3)[1−(−(1/2))^(n−4) ]+4=((14)/3)−((32)/3)(−(1/2))^n

MethodII―an+2−an+1=−12(an+1−an)letbn=an+1−anbn+1=−12bnbn=(−12)bn−1=(−12)2bn−2=...=(−12)n−4b4an+1−an=(−12)n−4(a5−a4)=(−12)n−4an+1−an=(−12)n−4=16(−12)nan−an−1=16(−12)n−1an−1−an−2=16(−12)n−2...a5−a4=16(−12)4an−a4=16(−12)4[1+(−12)+(−12)2+...+(−12)n−5]an−a4=1−(−12)n−41+12an=23[1−(−12)n−4]+4=143−323(−12)n

Answered by mr W last updated on 05/May/24

Method I 2a_(n+2) −a_(n+1) −a_n =0 2r^2 −r−1=0 (2r+1)(r−1)=0 r=−(1/2), 1 a_n =A(−(1/2))^n +B a_n =2a_(n+2) −a_(n+1) a_3 =2×5−4=6 a_2 =2×4−6=2 a_1 =2×6−2=10 a_0 =2×2−10=−6 −6=A+B 10=−(1/2)A+B ⇒A=−((32)/3) ⇒B=((14)/3) ⇒a_n =((14)/3)−((32)/3)(−(1/2))^n

MethodI―2an+2−an+1−an=02r2−r−1=0(2r+1)(r−1)=0r=−12,1an=A(−12)n+Ban=2an+2−an+1a3=2×5−4=6a2=2×4−6=2a1=2×6−2=10a0=2×2−10=−6−6=A+B10=−12A+B⇒A=−323⇒B=143⇒an=143−323(−12)n

Commented by efronzo1 last updated on 06/May/24

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