Question

Fie progresia geometrica in care b3=2 radical din 2 si b6=8 . Sa se calculeze suma primilor 10 termeni .

Answer 1

[tex]\frac{b_6}{b_3}=q^3=\sqrt2\Rightarrowq= \sqrt[6]{2}\Rightarrow \\ b_1=2\sqrt2:\sqrt[6]{2^2} \\ S_{10}=b_1+b_2+...+b_{10}=\frac{2\sqrt2}{\sqrt[6]{2^2}} \frac{\sqrt[6]{2}^{10}-1 }{\sqrt[6]{2} -1}[/tex]