f(n)=2\cdot(-3)^{\large{\,n}}

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:

\[
f(1) = -6
\]
\[
f(n) = -3 \cdot f(n-1) \text{ for } n > 1
\]