Question

What is the recursive formula for the geometric sequence with this explicit formula?

Answer 1

Answer:

$\large\huge\boxed{\left\{\begin{array}{ccc}a_1=9\\a_n=a_{n-1}\cdot\left(-\dfrac{1}{3}\right)\end{array}\right}$

Step-by-step explanation:

$a_n=9\cdot\left(-\dfrac{1}{3}\right)^{n-1}\\\\\text{Calculate}\ a_1.\ \text{Put n = 1 to the explicit formula of the geometric sequence:}\\\\a_1=9\cdot\left(-\dfrac{1}{3}\right)^{1-1}=9\cdot\left(-\dfraC{1}{3}\right)^0=9\cdot1=9\\\\\text{Calculate the common ratio:}\\\\r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=9\cdot\left(-\dfrac{1}{3}\right)^{n+1-1}=9\cdot\left(-\dfrac{1}{3}\right)^n$

$r=\dfrac{9\!\!\!\!\diagup^1\cdot\left(-\frac{1}{3}\right)^n}{9\!\!\!\!\diagup_1\cdot\left(-\frac{1}{3}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^m}{a^n}=a^{m-n}\\\\r=\left(-\dfrac{1}{3}\right)^{n-(n-1)}=\left(-\dfrac{1}{3}\right)^{n-n-(-1)}=\left(-\dfrac{1}{3}\right)^1=-\dfrac{1}{3}\\\\a_n=a_{n-1}\cdot\left(-\dfrac{1}{3}\right)$