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if-u-n-1-5-1-5-2-n-1-5-2-n-then-u-n-1-u-n-u-n-1-n-0-1-2-




Question Number 194579 by MM42 last updated on 10/Jul/23
if    u_n =(1/( (√5)))[(((1+(√5))/2))^n −(((1−(√5))/2))^n ]   then   u_(n+1) =u_n +u_(n−1)    ?     ;   n=0,1,2,..
ifun=15[(1+52)n(152)n]thenun+1=un+un1?;n=0,1,2,..
Answered by Frix last updated on 10/Jul/23
n={0, 1, 2, 3, 4, 5, 6, 7, ...}  u_n ={0, 1, 1, 2, 3, 5, 8, 13, ...}
n={0,1,2,3,4,5,6,7,}un={0,1,1,2,3,5,8,13,}
Commented by MM42 last updated on 10/Jul/23
fibonacci  sequence  please  prove...
fibonaccisequencepleaseprove
Answered by MM42 last updated on 13/Jul/23
prove  u_n +u_(n−1) =(1/( (√5)))[(((1+(√5))/2))^n −(((1−(√5))/2))^n +(((1+(√5))/2))^(n−1) −(((1−(√5))/2))^(n−1) ]  =(1/( (√5)))[(((1+(√5))/2))^(n−1) (((1+(√5))/2)+1)−(((1−(√5))/2))^(n−1) (((1−(√5))/2)+1)]  =(1/( (√5)))[(((1+(√5))/2))^(n−1) (((3+(√5))/2))−(((1−(√5))/2))^(n−1) (((3−(√5))/2))]  =(1/( (√5)))[(((1+(√5))/2))^n −(((1−(√5))/2))^n ]=u_(n+1)   metod 2  let  a=((1+(√5))/2)  &  b=((1−(√5))/2)  ⇒u_n =(1/( (√5)))(a^n −b^n )  a+b=1  &  ab=−1   ⇒ x^2 +x−1=0  ; a,b , are the roots   a+b=1⇒a^(n+1) +a^n b=a^n  ⇒a^(n+1) −a^(n−1) =a^n   (i)   b^(n+1) +ab^n =b^n  ⇒b^(n+1) −b^(n−1) =b^n   (ii)  (i)+(ii)⇒a^(n+1) −b^(n+1) =(a^n −b^n )+(a^(n−1) −b^(n−1) )  ⇒u_(n+1) =u_m +u_(n−1)
proveun+un1=15[(1+52)n(152)n+(1+52)n1(152)n1]=15[(1+52)n1(1+52+1)(152)n1(152+1)]=15[(1+52)n1(3+52)(152)n1(352)]=15[(1+52)n(152)n]=un+1metod2leta=1+52&b=152un=15(anbn)a+b=1&ab=1x2+x1=0;a,b,aretherootsa+b=1an+1+anb=anan+1an1=an(i)bn+1+abn=bnbn+1bn1=bn(ii)(i)+(ii)an+1bn+1=(anbn)+(an1bn1)un+1=um+un1

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