a) Dziesięć początkowych wyrazów ciągu Fibonacciego:

$a_1=1$  

$a_2=1$  

$a_3=a_1+a_2=1+1=2$  

$a_4=a_2+a_3=1+2=3$  

$a_5=a_3+a_4=2+3=5$  

$a_6=a_4+a_5=3+5=8$  

$a_7=a_5+a_6=5+8=13$  

$a_8=a_6+a_7=8+13=21$  

$a_9=a_7+a_8=13+21=34$  

$a_{10}=a_8+a_9=21+34=55$   

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b) Rozwiązanie przedstawiono na rysunku:

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c) Ze wzoru ogólnego ciągu Fibonacciego otrzymujemy:

$a_1=\frac{1}{\sqrt{5}}\left({\left(\frac{1+\sqrt{5}}{2}\right)}^1-{\left(-\frac{1}{\frac{1+\sqrt{5}}{2}}\right)}^1\right)=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}-\left(-\frac{2}{1+\sqrt{5}}\right)\right)=$  

$=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}+\frac{2}{1+\sqrt{5}}\right)=\frac{1}{\sqrt{5}}\cdot \frac{\left(1+\sqrt{5}\right)\left(1+\sqrt{5}\right)+2\cdot 2}{2\left(1+\sqrt{5}\right)}=$  

$=\frac{1}{\sqrt{5}}\cdot \frac{1+2\sqrt{5}+5+4}{2\left(1+\sqrt{5}\right)}=\frac{1}{\sqrt{5}}\cdot \frac{10+2\sqrt{5}}{2\left(1+\sqrt{5}\right)}=\frac{1}{\sqrt{5}}\cdot \frac{2\left(5+\sqrt{5}\right)}{2\left(1+\sqrt{5}\right)}=$  

$=\frac{1}{\sqrt{5}}\cdot \frac{5+\sqrt{5}}{1+\sqrt{5}}=\frac{5+\sqrt{5}}{\sqrt{5}\left(1+\sqrt{5}\right)}=\frac{5+\sqrt{5}}{\sqrt{5}+5}=\frac{5+\sqrt{5}}{5+\sqrt{5}}=1$  

$a_2=\frac{1}{\sqrt{5}}\left({\left(\frac{1+\sqrt{5}}{2}\right)}^2-{\left(-\frac{1}{\frac{1+\sqrt{5}}{2}}\right)}^2\right)=\frac{1}{\sqrt{5}}\left(\frac{{\left(1+\sqrt{5}\right)}^2}{2^2}-{\left(-\frac{2}{1+\sqrt{5}}\right)}^2\right)=$  

$=\frac{1}{\sqrt{5}}\left(\frac{1+2\sqrt{5}+5}{4}-\frac{{\left(-2\right)}^2}{{\left(1+\sqrt{5}\right)}^2}\right)=\frac{1}{\sqrt{5}}\left(\frac{6+2\sqrt{5}}{4}-\frac{4}{1+2\sqrt{5}+5}\right)=$  

$=\frac{1}{\sqrt{5}}\left(\frac{2\left(3+\sqrt{5}\right)}{4}-\frac{4}{6+2\sqrt{5}}\right)=\frac{1}{\sqrt{5}}\left(\frac{3+\sqrt{5}}{2}-\frac{4}{2\left(3+\sqrt{5}\right)}\right)=$  

$=\frac{1}{\sqrt{5}}\left(\frac{3+\sqrt{5}}{2}-\frac{2}{3+\sqrt{5}}\right)=\frac{1}{\sqrt{5}}\cdot \frac{\left(3+\sqrt{5}\right)\left(3+\sqrt{5}\right)-2\cdot 2}{2\left(3+\sqrt{5}\right)}=$  

$=\frac{1}{\sqrt{5}}\cdot \frac{9+6\sqrt{5}+5-4}{2\left(3+\sqrt{5}\right)}=\frac{1}{\sqrt{5}}\cdot \frac{10+6\sqrt{5}}{2\left(3+\sqrt{5}\right)}=\frac{1}{\sqrt{5}}\cdot \frac{2\left(5+3\sqrt{5}\right)}{2\left(3+\sqrt{5}\right)}=$  

$=\frac{1}{\sqrt{5}}\cdot \frac{5+3\sqrt{5}}{3+\sqrt{5}}=\frac{5+3\sqrt{5}}{\sqrt{5}\left(3+\sqrt{5}\right)}=\frac{5+3\sqrt{5}}{3\sqrt{5}+5}=\frac{5+3\sqrt{5}}{5+3\sqrt{5}}=1$  

Korzystając z kalkulatora obliczamy:

$a_3=\frac{1}{\sqrt{5}}\left({\left(\frac{1+\sqrt{5}}{2}\right)}^3-{\left(-\frac{1}{\frac{1+\sqrt{5}}{2}}\right)}^3\right)=2$  

$a_4=\frac{1}{\sqrt{5}}\left({\left(\frac{1+\sqrt{5}}{2}\right)}^4-{\left(-\frac{1}{\frac{1+\sqrt{5}}{2}}\right)}^4\right)=3$