To derive the recursive formula for \( f(n) = 2 \cdot (-3)^n \), we first need to determine \( f(1) \):
\[ f(1) = 2 \cdot (-3)^1 = 2 \cdot (-3) = -6 \]
Next, to find the recursive relationship, we express \( f(n) \) in terms of \( f(n-1) \).
Starting with the expression for \( f(n) \):
\[ f(n) = 2 \cdot (-3)^n \]
We can relate \( f(n) \) to \( f(n-1) \):
\[ f(n-1) = 2 \cdot (-3)^{n-1} \]
Now, we can express \( f(n) \) in terms of \( f(n-1) \):
\[ f(n) = 2 \cdot (-3)^n = 2 \cdot (-3) \cdot (-3)^{n-1} = -3 \cdot (2 \cdot (-3)^{n-1}) = -3 \cdot f(n-1) \]
Now we can complete the recursive formula:
\[ f(1) = -6 \] \[ f(n) = -3 \cdot f(n-1) \text{ for } n > 1 \]
Putting it all together, the completed recursive formulas are:
\[ f(1) = -6 \] \[ f(n) = -3 \cdot f(n-1) \text{ for } n > 1 \]