Question

va rog frumos !!!!
dau coroana​

Answer 1

Răspuns:

Explicație pas cu pas:

termenul general al unei progresii geometrice:

$b_{n} =b_{1} \cdot q^{n-1}$

$a) b_{1} =6,q=18:6=3\\\ \\ \ b_{n} =6\cdot 3^{n-1} \\ \\ b) b_{1} =2,q=1:2=\frac{1}{2} \\\\ b_{n} =2\cdot (\frac{1}{2}) ^{n-1} =\frac{1}{2^{n-2} } \\ \\ c)b_{1} =-1,q=\frac{1}{3} :(-1)=-\frac{1}{3} \\\\ b_{n} =-1\cdot (-\frac{1}{3})^{n-1} =\frac{(-1)^{n-2} }{3^{n-1} } \\ \\ d)b_{1} =\sqrt{2} ,q=1:\sqrt{2} =\frac{1}{\sqrt{2} } =\frac{\sqrt{2} }{2} \\ \\ b_{n} =\sqrt{2} \cdot (\frac{\sqrt{2} }{2})^{n-1} =\frac{\sqrt{2} ^{n} }{2^{n-1} } \\ \\ e)b_{1} =\sqrt{6} ,q=\sqrt{3} :\sqrt{6} =\frac{1}{\sqrt{2} } =$

$=\frac{\sqrt{2} }{2} \\ \\ b_{n} =\sqrt{6} \cdot (\frac{\sqrt{2} }{2} )^{n-1}$