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- Tinku Tara July 10, 2023 Algebra 0 Comments FacebookTweetPin Question Number 194579 by MM42 last updated on 10/Jul/23 ifun=15[(1+52)n−(1−52)n]thenun+1=un+un−1?;n=0,1,2,.. Answered by Frix last updated on 10/Jul/23 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 fibonaccisequencepleaseprove… Answered by MM42 last updated on 13/Jul/23 proveun+un−1=15[(1+52)n−(1−52)n+(1+52)n−1−(1−52)n−1]=15[(1+52)n−1(1+52+1)−(1−52)n−1(1−52+1)]=15[(1+52)n−1(3+52)−(1−52)n−1(3−52)]=15[(1+52)n−(1−52)n]=un+1metod2leta=1+52&b=1−52⇒un=15(an−bn)a+b=1&ab=−1⇒x2+x−1=0;a,b,aretherootsa+b=1⇒an+1+anb=an⇒an+1−an−1=an(i)bn+1+abn=bn⇒bn+1−bn−1=bn(ii)(i)+(ii)⇒an+1−bn+1=(an−bn)+(an−1−bn−1)⇒un+1=um+un−1 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-194573Next Next post: abc-e-3-d-3-f-3-edf-a-3-b-3-c-3-find-abc-and-edf- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.