Beregn
[tex]\sum_{n=1}^{\infty} \frac{F_n}{3^n}[/tex]
der [tex]F_n[/tex] er det n-te Fibonacci-tallet ([tex]F_1=F_2=1\, ,\,F_{n+2}=F_{n+1}+F_n[/tex])
Fibonacci
Moderatorer: Vektormannen, espen180, Aleks855, Solar Plexsus, Gustav, Nebuchadnezzar, Janhaa
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F[sub]n[/sub] på lukket form:
[tex]F_n=\frac{\varphi^n-(1-\varphi)^n}{\sqrt 5}=\frac{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(1-\frac{1+\sqrt{5}}{2}\right)^n}{\sqrt{5}}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right) \\ F_n=\frac{1}{\sqrt{5}}\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n} \\ \frac{F_n}{3^n}=\frac{1}{\sqrt{5}}\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n3^n}=\frac{1}{\sqrt{5}}\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{6^n}=\frac{1}{\sqrt{5}}\left(\left(\frac{(1+\sqrt{5})}{6}\right)^n-\left(\frac{(1-\sqrt{5})}{6}\right)^n\right) \\ S=\sum_{n=0}^\infty \frac{F_n}{3^n}=\frac{1}{\sqrt{5}}\sum_{n=0}^\infty \left(\frac{(1+\sqrt{5})}{6}\right)^n - \frac{1}{\sqrt{5}}\sum_{n=0}^\infty \left(\frac{(1-\sqrt{5})}{6}\right)^n=\frac{1}{\sqrt{5}}\left(\frac{1}{1-\frac{(1+\sqrt{5})}{6}}-\frac{1}{1-\frac{(1-\sqrt{5})}{6}\right) \\ S=\frac{1}{\sqrt{5}}\left(\frac{6}{5-\sqrt{5}}-\frac{6}{5+\sqrt{5}}\right)=\frac{1}{\sqrt{5}}\left(\frac{6(5+\sqrt{5})}{25-5}-\frac{6(5-\sqrt{5})}{25-5}\right)=\frac{30+6\sqrt{5}-30+6\sqrt{5}}{20\sqrt{5}}=\frac{3}{5} \\ \sum_{n=0}^{\infty} \frac{F_n}{3^n}=\frac35[/tex]
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F[sub]n[/sub] på lukket form:
[tex]F_n=\frac{\varphi^n-(1-\varphi)^n}{\sqrt 5}=\frac{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(1-\frac{1+\sqrt{5}}{2}\right)^n}{\sqrt{5}}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right) \\ F_n=\frac{1}{\sqrt{5}}\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n} \\ \frac{F_n}{3^n}=\frac{1}{\sqrt{5}}\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n3^n}=\frac{1}{\sqrt{5}}\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{6^n}=\frac{1}{\sqrt{5}}\left(\left(\frac{(1+\sqrt{5})}{6}\right)^n-\left(\frac{(1-\sqrt{5})}{6}\right)^n\right) \\ S=\sum_{n=0}^\infty \frac{F_n}{3^n}=\frac{1}{\sqrt{5}}\sum_{n=0}^\infty \left(\frac{(1+\sqrt{5})}{6}\right)^n - \frac{1}{\sqrt{5}}\sum_{n=0}^\infty \left(\frac{(1-\sqrt{5})}{6}\right)^n=\frac{1}{\sqrt{5}}\left(\frac{1}{1-\frac{(1+\sqrt{5})}{6}}-\frac{1}{1-\frac{(1-\sqrt{5})}{6}\right) \\ S=\frac{1}{\sqrt{5}}\left(\frac{6}{5-\sqrt{5}}-\frac{6}{5+\sqrt{5}}\right)=\frac{1}{\sqrt{5}}\left(\frac{6(5+\sqrt{5})}{25-5}-\frac{6(5-\sqrt{5})}{25-5}\right)=\frac{30+6\sqrt{5}-30+6\sqrt{5}}{20\sqrt{5}}=\frac{3}{5} \\ \sum_{n=0}^{\infty} \frac{F_n}{3^n}=\frac35[/tex]
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Sist redigert av espen180 den 30/12-2009 14:05, redigert 1 gang totalt.
Alternativt:
[tex]\sum^\infty_{n=1} \frac{F_n}{3^n}=\frac49 +\sum^\infty_{n=3} \frac{F_{n-1}+F_{n-2}}{3^n}=\frac49 + \frac{1}{3}\sum^\infty_{n=2} \frac{F_n}{3^n}+\frac{1}{9}\sum^\infty_{n=1} \frac{F_n}{3^n} \\ =\frac13 + (\frac{1}{3}+\frac{1}{9})\sum^\infty_{n=1}\frac{F_n}{3^n} \Rightarrow \sum^\infty_{n=1}\frac{F_n}{3^n}=\frac{\frac{1}{3}}{1-\frac{4}{9}} =\frac{3}{5}[/tex]
[tex]\sum^\infty_{n=1} \frac{F_n}{3^n}=\frac49 +\sum^\infty_{n=3} \frac{F_{n-1}+F_{n-2}}{3^n}=\frac49 + \frac{1}{3}\sum^\infty_{n=2} \frac{F_n}{3^n}+\frac{1}{9}\sum^\infty_{n=1} \frac{F_n}{3^n} \\ =\frac13 + (\frac{1}{3}+\frac{1}{9})\sum^\infty_{n=1}\frac{F_n}{3^n} \Rightarrow \sum^\infty_{n=1}\frac{F_n}{3^n}=\frac{\frac{1}{3}}{1-\frac{4}{9}} =\frac{3}{5}[/tex]
Ikke at jeg er kompetent nok til å skjønne noe som helst av dette her, men det første jeg kastet blikket mitt på her var dette:
Blingser jeg nå, eller er det noe som skurrer?espen180 skrev:[tex]\frac{30+6\sqrt{5}-30+6\sqrt{5}}{20\sqrt{5}}=\frac{3}{3}[/tex]