Korzystamy ze wzoru:

$S=\frac{a_1}{1-q}\text{dla}q\in \left(-1,1\right)$  

 

$\text{a)}$  

$S=\frac{2}{1-\frac{\sqrt{3}}{3}}=\frac{2}{\frac{3}{3}-\frac{\sqrt{3}}{3}}=2\cdot \frac{3}{3-\sqrt{3}}=\frac{6}{3-\sqrt{3}}=$     

$=\frac{6\left(3+\sqrt{3}\right)}{3^2-{\sqrt{3}}^2}=\frac{6\left(3+\sqrt{3}\right)}{9-3}=\frac{6\left(3+\sqrt{3}\right)}{6}=3+\sqrt{3}$  

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$\text{b)}$  

$a_1+a_2=-12$  

$-18+a_2=-12\mid +18$  

$a_2=6$  

 

$q=\frac{a_2}{a_1}=\frac{6}{-18}=-\frac{1}{3}$  

 

$S=\frac{-18}{1-\left(-\frac{1}{3}\right)}=\frac{-18}{\frac{4}{3}}=-18\cdot \frac{3}{4}=-\frac{54}{4}=-\frac{27}{2}$  

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$\text{c)}$  

$\{\begin{array}{l}{a}_1+a_3=10\\ a_2+a_4=-5\end{array}$  

$\{\begin{array}{l}{a}_1+a_1{q}^2=10\\ a_{1}q+a_1{q}^3=-5\end{array}$    

$\{\begin{array}{l}{a}_1\left(1+q^2\right)=10\mid :a_1\\ a_{1}q\left(1+q^2\right)=-5\end{array}$  

$\{\begin{array}{l}1+q^2=\frac{10}{a_1}\\ a_{1}q\cdot \frac{10}{a_1}=-5\end{array}$  

Rozwiązując drugie równanie otrzymujemy:

$10q=-5\mid :10$  

$q=-\frac{1}{2}$  

Rozwiązując pierwsze równanie otrzymujemy:

$1+{\left(-\frac{1}{2}\right)}^2=\frac{10}{a_1}$  

$1+\frac{1}{4}=\frac{10}{a_1}$   

$\frac{5}{4}=\frac{10}{a_1}$  

$5a_1=40$  

$a_1=8$  

 

$S=\frac{8}{1-\left(-\frac{1}{2}\right)}=\frac{8}{\frac{3}{2}}=8\cdot \frac{2}{3}=\frac{16}{3}$  

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$\text{d)}$  

$\{\begin{array}{l}{a}_3=\sqrt{2}\\ 4a_6={\left(a_3\right)}^2\end{array}$  

$\{\begin{array}{l}{a}_1\cdot q^2=\sqrt{2}\\ 4a_1\cdot q^5={\sqrt{2}}^2\end{array}$  

$\{\begin{array}{l}{a}_1=\frac{\sqrt{2}}{q^2}\\ 4\cdot \frac{\sqrt{2}}{q^2}\cdot q^5=2\end{array}$  

Rozwiązując drugą równość otrzymujemy:

$4\sqrt{2}\cdot q^3=2\mid :4\sqrt{2}$  

$q^3=\frac{1}{2\sqrt{2}}$  

$q^3={\left(2^1\cdot 2^{\frac{1}{2}}\right)}^{-1}$  

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

$q^3={\left(2^{\frac{3}{2}}\right)}^{-1}$  

$q^3=2^{-\frac{3}{2}}$  

$q^3={\left(2^{-\frac{1}{2}}\right)}^3$  

$q=2^{-\frac{1}{2}}$  

$q=\sqrt{\frac{1}{2}}$  

$q=\frac{\sqrt{2}}{2}$  

 

$a_1=\frac{\sqrt{2}}{\frac{1}{2}}$  

$a_1=2\sqrt{2}$  

 

$S=\frac{2\sqrt{2}}{1-\frac{\sqrt{2}}{2}}=\frac{2\sqrt{2}}{\frac{2}{2}-\frac{\sqrt{2}}{2}}=2\sqrt{2}\cdot \frac{2}{2-\sqrt{2}}=\frac{4\sqrt{2}}{2-\sqrt{2}}=$  

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

$=4\sqrt{2}+4=4\left(\sqrt{2}+1\right)$