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$f(n)=2*(-3)^(n) Complete the recursive formula of f(n). {:[f(1)=◻],[f(n)=f(n-1)]:}$

$f(n)=2 \cdot(-3)^{n}$ $\newline$ Complete the recursive formula of $f(n)$ . $\newline$ $\begin{array}{l} f(1)=\square \\ f(n)=f(n-1) \cdot \square \end{array}$

Full solution

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

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Complete the recursive formula of

f(n)

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\begin{array}{l} f(1)=\square \\ f(n)=f(n-1) \cdot \square \end{array}

Given Explicit Formula: We are given the explicit formula for the sequence: $f(n) = 2 \times (-3)^n$ . To find the recursive formula, we need to express $f(n)$ in terms of $f(n-1)$ .
Find $f(1)$ : First, let's find $f(1)$ by substituting $n = 1$ into the explicit formula. $\newline$ $f(1) = 2 \times (-3)^{1} = 2 \times (-3) = -6.$ $\newline$ So, $f(1) = -6.$
Relationship between $f(n)$ and $f(n-1)$ : Now, let's find the relationship between $f(n)$ and $f(n-1)$ . We know that $f(n) = 2 \times (-3)^{n}$ and $f(n-1) = 2 \times (-3)^{n-1}$ .
Divide $f(n)$ by $f(n-1)$ : To find $f(n)$ in terms of $f(n-1)$ , we can divide $f(n)$ by $f(n-1)$ : $\frac{f(n)}{f(n-1)} = \frac{2 \cdot (-3)^n}{2 \cdot (-3)^{n-1}}.$
Simplify the Expression: Simplify the expression by canceling out common factors and using the properties of exponents: $\frac{f(n)}{f(n-1)} = \frac{(-3)^{n}}{(-3)^{n-1}} = (-3)^{n - (n - 1)} = (-3)^{1} = -3$ .
Express $f(n)$ in terms of $f(n-1)$ : Now we can express $f(n)$ in terms of $f(n-1)$ : $f(n) = -3 \times f(n-1)$ .
Combine Initial Condition: Combine the initial condition $f(1) = -6$ with the recursive relationship we just found: $f(n) = -3 \times f(n-1)$ , for n > 1.