Question

what is the recursive definition for this sequence? -4, -3, -2, -1, ... \

$$\begin{cases} f(0) = ? \\\\ f(n) = \\quadf(n - 1) + \\quad \\end{cases}$$

Explanation:

Step1: Find the initial term \( f(0) \)

The sequence starts with -4 when \( n = 0 \) (assuming \( f(0) \) corresponds to the first term). So \( f(0)=-4 \).

Step2: Analyze the pattern for \( f(n) \)

Looking at the sequence: -4, -3, -2, -1,... Each term is obtained by adding 1 to the previous term. So the recursive formula should be \( f(n)=1\times f(n - 1)+1 \), because multiplying by 1 leaves \( f(n - 1) \) unchanged and then adding 1 gives the next term.

Answer:

\( f(0) = -4 \)
\( f(n) = 1\times f(n - 1) + 1 \) (or simplified as \( f(n) = f(n - 1) + 1 \))