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Question Number 194693 by MM42 last updated on 13/Jul/23

if   f_n =f_(n−1) +f_(n−2)   ;  f_1 =f_2 =1  then   prove that   5∣f_(5n)

$${if}\:\:\:{f}_{{n}} ={f}_{{n}−\mathrm{1}} +{f}_{{n}−\mathrm{2}} \:\:;\:\:{f}_{\mathrm{1}} ={f}_{\mathrm{2}} =\mathrm{1} \\ $$$${then}\:\:\:{prove}\:{that}\:\:\:\mathrm{5}\mid{f}_{\mathrm{5}{n}} \:\: \\ $$

Answered by Frix last updated on 13/Jul/23

f_1 =k_1   f_2 =k_2   f_n =f_(n−1) +f_(n−2) ∀n≥3  ⇒  f_n =3f_(n−5) +5f_(n−4) ∀n≥6  5∣f_(n−5)  ⇔ 5∣f_n ∀n≥6

$${f}_{\mathrm{1}} ={k}_{\mathrm{1}} \\ $$$${f}_{\mathrm{2}} ={k}_{\mathrm{2}} \\ $$$${f}_{{n}} ={f}_{{n}−\mathrm{1}} +{f}_{{n}−\mathrm{2}} \forall{n}\geqslant\mathrm{3} \\ $$$$\Rightarrow \\ $$$${f}_{{n}} =\mathrm{3}{f}_{{n}−\mathrm{5}} +\mathrm{5}{f}_{{n}−\mathrm{4}} \forall{n}\geqslant\mathrm{6} \\ $$$$\mathrm{5}\mid{f}_{{n}−\mathrm{5}} \:\Leftrightarrow\:\mathrm{5}\mid{f}_{{n}} \forall{n}\geqslant\mathrm{6} \\ $$

Commented by MM42 last updated on 13/Jul/23

that′s right

$${that}'{s}\:{right} \\ $$

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