\begin{cases}

f(1)=-3
\\\\
f(n)=2 \cdot f(n-1) + 1
\end{cases}
f(2)=

1 answer

To find \( f(2) \), we can use the recursive formula given in the problem:

\[ f(n) = 2 \cdot f(n-1) + 1 \]

We also know the base case:

\[ f(1) = -3 \]

Now, we can calculate \( f(2) \):

\[ f(2) = 2 \cdot f(1) + 1 \]

Substituting the value of \( f(1) \):

\[ f(2) = 2 \cdot (-3) + 1 \]

Calculating this gives:

\[ f(2) = -6 + 1 = -5 \]

Therefore, the value of \( f(2) \) is:

\[ \boxed{-5} \]