Będziemy korzystać ze wzoru

$S_n=\{\begin{array}{l}{a}_1\cdot \frac{1-q^n}{1-q}\text{,}\text{dla }q\ne 1\\ a_1\cdot n\text{,}\text{ dla }q=1\end{array}$

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a) Mamy

$a_1=\frac{1}{4}$

$q=-2$

Stąd

$S_7=\frac{1}{4}\cdot \frac{1-{\left(-2\right)}^7}{1-\left(-2\right)}=\frac{1}{4}\cdot \frac{1+128}{3}==\frac{129}{12}=10\frac{3}{4}$    

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b) Mamy

$a_6=8\sqrt{2}$

$q=\sqrt{2}$

Zatem

$a_1=\frac{a_6}{q^5}=\frac{8\sqrt{2}}{4\sqrt{2}}=2$  

Stąd

$S_7=2\cdot \frac{1-{\left(\sqrt{2}\right)}^7}{1-\sqrt{2}}=2\cdot \frac{1-8\sqrt{2}}{1-\sqrt{2}}=2\cdot \frac{\left(1-8\sqrt{2}\right)\left(1+\sqrt{2}\right)}{1-2}=-2\left(1-8\sqrt{2}+\sqrt{2}-16\right)=\left(-2\right)\cdot \left(-7\sqrt{2}-15\right)=14\sqrt{2}+30$    

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c) Mamy

$a_1=-1$

$a_4=3\sqrt{3}$  

Zatem

$q^3=\frac{a_4}{a_1}=\frac{3\sqrt{3}}{-1}=-3\sqrt{3}$

$q=-\sqrt{3}$  

Stąd

$S_7=\left(-1\right)\cdot \frac{1-{\left(-\sqrt{3}\right)}^7}{1-\left(-\sqrt{3}\right)}=-\frac{1+27\sqrt{3}}{1+\sqrt{3}}=-\frac{\left(1+27\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}=\frac{1+27\sqrt{3}-\sqrt{3}-81}{2}=\frac{26\sqrt{3}-80}{2}=13\sqrt{3}-40$