Question

what is the recursive definition for this sequence? 2, 12, 72, 432, ... \

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

Explanation:

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

The first term of the sequence is \( 2 \), so when \( n = 0 \), \( f(0)=2 \).

Step2: Analyze the pattern between terms

Check the ratio between consecutive terms: \( \frac{12}{2} = 6 \), \( \frac{72}{12}=6 \), \( \frac{432}{72}=6 \). So each term is \( 6 \) times the previous term. This means the recursive formula should be \( f(n)=6f(n - 1)+0 \) (since there's no additional constant, just multiplication by 6).

Answer:

The recursive definition is \(

$$\begin{cases} f(0) = 2 \\ f(n) = 6f(n - 1)+0 \end{cases}$$

\)

So \( f(0)=\boldsymbol{2} \), the coefficient of \( f(n - 1) \) is \( \boldsymbol{6} \), and the constant term is \( \boldsymbol{0} \).