Question and Answers Forum

All Questions      Topic List

Set Theory Questions

Previous in All Question      Next in All Question      

Previous in Set Theory      Next in Set Theory      

Question Number 194709 by MM42 last updated on 13/Jul/23

$$Showthatinfibonaccisequence$$$$f_{3n}=f_n^3+f_{n+1}^3-f_{n-1}^3$$$$$$

Answered by TheHoneyCat last updated on 14/Jul/23

$$\mathrm{Let}\phi :=\frac{1+\sqrt{5}}{2}\mathrm{and}\psi :=\frac{1-\sqrt{5}}{2}$$$$\mathrm{You}\mathrm{might}\mathrm{know}\mathrm{that}{f}_n=\frac{{\phi }^n-{\psi }^n}{\sqrt{5}}$$$$\mathrm{from}\mathrm{that}\mathrm{we}\mathrm{can}\mathrm{compute}:$$$$X=f_n^3+f_{n+1}^3-f_{n-1}^3$$$$=\frac{1}{\sqrt{5}}\times \frac{1}{5}({({\phi }^n-{\psi }^n)}^3+{({\phi }^{n+1}-{\psi }^{n+1})}^3-{({\phi }^{n-1}-{\psi }^{n-1})}^3)$$$$=\frac{1}{\sqrt{5}}\times \frac{1}{5}({\phi }^{3n}-3{\phi }^{2n}{\psi }^n+3{\phi }^n{\psi }^{2n}-{\psi }^{3n}$$$$+{\phi }^{3n+3}-3{\phi }^{2n+2}{\psi }^{n+1}+3{\phi }^{n+1}{\psi }^{2n+2}-{\psi }^{3n+3}$$$$-{\phi }^{3n-3}+3{\phi }^{2n-2}{\psi }^{n-1}-3{\phi }^{n-1}{\psi }^{2n-2}+{\psi }^{3n-3})$$$$=\frac{1}{5}(f_{3n}+f_{3n+3}-f_{3n-3})+\frac{3}{5\times \sqrt{5}}(-{\phi }^{2n}{\psi }^n+{\phi }^n{\psi }^{2n}$$$$-{\phi }^{2n+2}{\psi }^{n+1}+{\phi }^{n+1}{\psi }^{2n+2}+{\phi }^{2n-2}{\psi }^{n-1}-{\phi }^{n-1}{\psi }^{2n-2})$$$$=\frac{1}{5}(f_{3n}+f_{3n+1}+f_{3n+2}-f_{3n-3})$$$$+\frac{3{\phi }^{n-1}{\psi }^{n-1}}{5\times \sqrt{5}}(-{\phi }^{n+1}\psi +\phi {\psi }^{n+1}-{\phi }^{n+3}{\psi }^2$$$$+{\phi }^2{\psi }^{n+3}+{\phi }^{n-1}-{\psi }^{n-1})$$$$=\frac{1}{5}(f_{3n}+f_{3n}+f_{3n-1}+f_{3n+1}+f_{3n}-f_{3n-3})$$$$+\frac{3{\phi }^{n-1}{\psi }^{n-1}}{5\times \sqrt{5}}(-{\phi }^{n+1}\psi +\phi {\psi }^{n+1}-{\phi }^{n+3}{\psi }^2$$$$+{\phi }^2{\psi }^{n+3}+{\phi }^{n-1}-{\psi }^{n-1})$$$$$$$$\mathrm{Now}\mathrm{this}\mathrm{is}\mathrm{the}\mathrm{moment}\mathrm{where}\mathrm{you}\mathrm{use}\mathrm{the}\mathrm{fact}$$$$\mathrm{that}\psi =\frac{-1}{\phi }$$$$X=\frac{1}{5}(3f_{3n}+f_{3n-1}+f_{3n}+f_{3n-1}-f_{3n-3})$$$$+\frac{3{(-1)}^{n+1}}{5\sqrt{5}}({\phi }^n-{\psi }^n-{\phi }^{n+1}+{\psi }^{n-1}+{\phi }^{n-1}-{\psi }^{n-1})$$$$=\frac{1}{5}(4f_{3n}+f_{3n-1}+f_{3n-2}+f_{3n-3}-f_{3n-3})$$$$+\frac{3{(-1)}^{n+1}}{5}(f_{n-1}+f_n-f_{n+1})$$$$=\frac{1}{5}(4f_{3n}+f_{3n-1}+f_{3n-2})+0$$$$=\frac{1}{5}(4f_{3n}+f_{3n})$$$$=f_{3n}{}_{◻}$$$$$$$$\mathrm{There}\mathrm{you}\mathrm{go}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com