Answer:
[tex]a_{n}=\frac{1}{4}a_{n-1}\\a_{1}=2[/tex]
Step-by-step explanation:
[tex]a_{n1}=2,\ a_{2}=\frac{1}{2},\ a_{3}=\frac{1}{8},\ a_{4}=\frac{1}{32}\\\\\frac{a_{2}}{a_{1}}=\frac{\frac{1}{2}}{2}=\frac{1}{4}\\\\\frac{a_{3}}{a_{2}}=\frac{(\frac{1}{8})}{(\frac{1}{2})}=\frac{1}{4}\\\\\frac{a_{4}}{a_{3}}=\frac{(\frac{1}{32})}{(\frac{1}{8})}=\frac{1}{4}[/tex]
So
this is a geometric series with first term [tex]=2[/tex]
and constant rate [tex]\frac{1}{4}[/tex]
[tex]n^{th}[/tex] term is given by [tex]a_{n}=2(\frac{1}{4})^{n-1}[/tex]
[tex](n-1)^{th}\ term=2(\frac{1}{4})^{n-1-1}=2(\frac{1}{4})^{n-2}\\\\\frac{a_n}{a_{n-1}}=\frac{2(\frac{1}{4})^{n-1}}{2(\frac{1}{4})^{n-2}}}=\frac{\frac{1}{4}(\frac{1}{4})^{n-2}}{(\frac{1}{4})^{n-2}}}=\frac{1}{4}\\\\\frac{a_{n}}{a_{n-1}}=\frac{1}{4}\\\\a_n=\frac{1}{4}a_{n-1}[/tex]
