tutaj<\/a>.<\/p>\n\\[F_n=\\dfrac{\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n-\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n}{\\sqrt{5}}.\\]<\/p>\n
\nNa pocz\u0105tek zauwa\u017cmy, \u017ce pierwiastkami r\u00f3wnania \\[\\begin{equation}x^2=x+1\\end{equation}\\]<\/p>\n
s\u0105 liczby \\(\\varphi\\) oraz \\(\\tilde{\\phi}=\\dfrac{1-\\sqrt{5}}{2}\\).<\/p>\n
Najpierw poka\u017cemy, \u017ce dla \\(x\\in\\{\\varphi, \\tilde{\\phi}\\}\\) oraz \\(n>1\\) prawdziwa jest r\u00f3wno\u015b\u0107<\/p>\n
\\[x^n=F_nx+F_{n-1}.\\]<\/p>\n
Istotnie, dla \\(n=2\\) jest oczywi\u015bcie spe\u0142niona, gdy\u017c jest to po prostu r\u00f3wnanie \\(x^2=x+1\\). Teraz za\u0142\u00f3\u017cmy, \u017ce jest prawdziwa dla pewnego \\(k\\geq 1\\), tj.<\/p>\n
\\[x^k=F_kx+F_{k-1}. \\]<\/p>\n
W\u00f3wczas<\/p>\n
\\[x^{k+1}=x^k\\cdot x=F_kx^2+F_{k-1}x=F_k(x+1)+F_{k-1}x=F_{k+1}x+ F_k.\\]<\/p>\n
Na mocy zasady indukcji zupe\u0142nej, otrzymujemy \u017ce jest to prawda dla ka\u017cdego \\(n>1\\). Poniewa\u017c \\(\\varphi\\) oraz \\(\\tilde{\\phi}\\) s\u0105 pierwiastkami r\u00f3wnania \\[x^2=x+1,\\] to s\u0105 r\u00f3wnie\u017c pierwiastkami r\u00f3wnania<\/p>\n
\\[x^n=F_nx+F_{n-1}\\]<\/p>\n
gdy\u017c \\(F_nx+F_{n-1}=x^{n-2}\\cdot (x+1)\\). Mamy wi\u0119c<\/p>\n
\\[\\varphi^n=F_n\\varphi+F_{n-1} \\textrm{ oraz } \\tilde{\\phi}^n=F_n\\tilde{\\phi}+F_{n-1}.\\]<\/p>\n
Odejmuj\u0105c stronami, otrzymujemy<\/p>\n
\\[\\varphi^n-\\tilde{\\phi}^n=F_n\\cdot (\\varphi-\\tilde{\\phi}).\\]<\/p>\n
Czyli<\/p>\n
\\[\\left(\\dfrac{1+\\sqrt{5}}{2}\\right)^n-\\left(\\dfrac{1-\\sqrt{5}}{2}\\right)^n=F_n\\cdot\\left(\\dfrac{1+\\sqrt{5}}{2}-\\dfrac{1-\\sqrt{5}}{2}\\right)=F_n\\cdot\\sqrt{5}.\\]<\/p>\n
Dziel\u0105c przez \\(\\sqrt{5}\\) otrzymujemy wz\u00f3r Bineta.<\/p>\n","protected":false},"excerpt":{"rendered":"
Celem tego wpisu jest wyprowadzenie wzoru Bineta na n-t\u0105 liczb\u0119 ci\u0105gu Fibonacciego.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7,5],"tags":[8,13],"yoast_head":"\n
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