Using induction, prove that F 2n =F n L n What I have so far: Assuming F 2k =F k

Question

Using induction, prove that   F2n =FnLn What I have so far: Assuming   F2k =FkLk   (then it would remain toshow that   F2k+2 =Fk+1Lk+1 ) I had already proven an earlier problem, by induction,that      F2k = F1 +F3 + F5 + ... + F2k-1 so F2k+2 = F1 + F3 +F5 + ... + F2k-1 + F2k+1 which, by assumtion is F2k+2 = FkLk +F2k+1 Am I on the right track? Any help would be much appreciated.Thanks. Using induction, prove that   F2n =FnLn What I have so far: Assuming   F2k =FkLk   (then it would remain toshow that   F2k+2 =Fk+1Lk+1 ) I had already proven an earlier problem, by induction,that      F2k = F1 +F3 + F5 + ... + F2k-1 so F2k+2 = F1 + F3 +F5 + ... + F2k-1 + F2k+1 which, by assumtion is F2k+2 = FkLk +F2k+1 Am I on the right track? Any help would be much appreciated.Thanks. Am I on the right track? Any help would be much appreciated.Thanks.

Explanation / Answer

I claim that if we know the following things to be true: Fn-1Ln + FnLn-1 =2F2n-1       (assumption1) FnLn =F2n                                     (assumption 2)Fn-1Ln-1 =F2n-2                    (assumption 3) ....then the following must also be true: FnLn+1 + Fn+1Ln =2F2n+1        (a) Fn+1Ln+1=F2n+2                        (b) FnLn=F2n                                               (c) If my claim is correct, then we can endlessly advance n by 1, andthis will complete the induction. Now, we prove my claim:     FnLn+1 +Fn+1Ln = Fn(Ln-1 + Ln) +Ln(Fn-1 +Fn)                (split Fn+1 into Fn-1 + Fn , samefor L) = FnLn-1 + FnLn +Fn-1Ln +FnLn              (expanded everything, combined 1st and 3rd term by using our firstassumption = 2F2n-1 +2F2n                                                        and then used our second assumption for the 2nd and 4th term) = 2F2n+1 So this proves (a). (c) is already truesince it's the same as assumption 2, now we just have to prove(b):      Fn+1Ln+1 = (Fn-1 + Fn)(Ln-1 +Ln)                                (split Fn+1 into Fn-1 + Fn , samefor L) = Fn-1Ln-1 + Fn-1Ln +FnLn-1 +FnLn           (expanded everything, applied assumptions 3 and 2 to 1st and 4thterms, = F2n-2 + 2F2n-1 +F2n                                   applied assumption1 to 2nd and 3rd terms) = [(F2n-2 + F2n-1) + F2n-1] +F2n                  (rearranged terms. Adding terms together in order of theparenthesis =F2n+2                                                                              lead to our final answer, and proves my claim to be correct)

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