Answer:
The recursive formulas is \(a_n=3\cdot a_{n-1}\).
Step-by-step explanation:
The geometric sequence is defined by the formula
\(a_n=2\cdot3^{n-1}\)
The first term of the sequence is
\(a_1=2\cdot3^{1-1}=2\cdot3^0=2\)
The second term of the sequence is
\(a_2=2\cdot3^{2-1}=2\cdot3=6\)
Similarly (n-1)th term is
\(a_{n-1}=2\cdot3^{n-1-1}=2\cdot3^{n-2}\)
The common ratio of a geometric sequence is
\(r=\frac{a_2}{a_1}=\frac{6}{2}=3\)
The recursive formulas for the geometric sequence is calculated as
\(a_n=2\cdot3^{n-1}\)
\(a_n=2\cdot3^{n-1-1}\cdot 3\) \([\because a^{m+n}=a^ma^n]\)
\(a_n=2\cdot3^{n-2}\cdot 3\)
\(a_n=a_{n-1}\cdot 3\) \([\because a_{n-1}=2\cdot3^{n-2}]\)
\(a_n=3\cdot a_{n-1}\)
Therefore the recursive formulas is \(a_n=3\cdot a_{n-1}\).
