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$f(n)=5*(-2)^(n-1) Complete the recursive formula of f(n). {:[f(1)=◻],[f(n)=f(n-1).]:}$

$f(n)=5 \cdot(-2)^{n-1}$ $\newline$ 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)=5 \cdot(-2)^{n-1}

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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 Sequence: We are given the explicit formula for a sequence: $f(n)=5\cdot(-2)^{n-1}$ . To find the recursive formula, we need to express $f(n)$ in terms of the previous term $f(n-1)$ .
Find $f(1)$ : First, let's find the value of the first term $f(1)$ by substituting $n=1$ into the explicit formula. $\newline$ $f(1) = 5 \cdot (-2)^{1-1} = 5 \cdot (-2)^0 = 5 \cdot 1 = 5$
Find $f(2)$ : Now, let's find the value of the second term $f(2)$ to understand the relationship between consecutive terms. $f(2) = 5 \cdot (-2)^{2-1} = 5 \cdot (-2)^1 = 5 \cdot (-2) = -10$
Identify Common Ratio: We can see that to go from $f(1)$ to $f(2)$ , we multiply $f(1)$ by $-2$ . This is the common ratio $r$ for the geometric sequence. So, the recursive formula will involve multiplying the previous term by $-2$ .
Recursive Formula: The recursive formula for the sequence can be written as: $\newline$ $f(n) = f(n-1) \times (-2)$ , for n > 1 $\newline$ And we already know that $f(1) = 5$ .
Complete Recursive Formula: Therefore, the complete recursive formula for the sequence is: ${f(1)=5}, {f(n)=f(n-1) \times (-2)}$ for n > 1