Dany jest ciąg:

$36,12\sqrt{6},24,\dots$  

a) Obliczmy iloraz tego ciągu:

$q=\frac{12\sqrt{6}}{36}=\frac{\sqrt{6}}{3}$  

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b) Zapiszmy wzór na n-ty wyraz tego ciągu:

$a_n=36\cdot {\left(\frac{\sqrt{6}}{3}\right)}^{n-1}=36\cdot {\left(\frac{\sqrt{6}}{3}\right)}^n\cdot {\left(\frac{\sqrt{6}}{3}\right)}^{-1}=$  

$=36\cdot {\left(\frac{\sqrt{6}}{3}\right)}^n\cdot \left(\frac{3}{\sqrt{6}}\right)=\frac{108}{\sqrt{6}}\cdot {\left(\frac{\sqrt{6}}{3}\right)}^n=$  

$=\frac{108\sqrt{6}}{6}\cdot {\left(\frac{\sqrt{6}}{3}\right)}^n=18\sqrt{6}\cdot {\left(\frac{\sqrt{6}}{3}\right)}^n$  

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c) Obliczmy sumę ośmiu początkowych wyrazów tego ciągu:

$S_8=36\cdot \frac{1-{\left(\frac{\sqrt{6}}{3}\right)}^8}{1-\frac{\sqrt{6}}{3}}=36\cdot \frac{1-\frac{6^4}{3^8}}{\frac{3-\sqrt{6}}{3}}=$  

$=36\cdot \frac{1-\frac{2^4\cdot 3^4}{3^8}}{\frac{3-\sqrt{6}}{3}}=36\cdot \left(1-\frac{2^4}{3^4}\right)\cdot \frac{3}{3-\sqrt{6}}=$  

$=36\cdot \left(1-\frac{16}{81}\right)\cdot \frac{3}{3-\sqrt{6}}=36\cdot \frac{65}{81}\cdot \frac{3}{3-\sqrt{6}}=4\cdot \frac{65}{9}\cdot \frac{3}{3-\sqrt{6}}=$  

$=4\cdot \frac{65}{3}\cdot \frac{1}{3-\sqrt{6}}=\frac{260}{3\cdot \left(3-\sqrt{6}\right)}=\frac{260\cdot \left(3+\sqrt{6}\right)}{3\cdot \left(9-6\right)}=$  

$=\frac{260\cdot \left(3+\sqrt{6}\right)}{3\cdot 3}=\frac{260\cdot \left(3+\sqrt{6}\right)}{9}$