Full solution

f(n)=45 \cdot\left(\frac{4}{5}\right)^{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 Explicit Formula: We are given the explicit formula for the sequence: $\newline$ $f(n) = 45 \times \left(\frac{4}{5}\right)^{n-1}$ $\newline$ 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: $\newline$ $f(1) = 45 \times \left(\frac{4}{5}\right)^{1-1} = 45 \times \left(\frac{4}{5}\right)^0 = 45 \times 1 = 45$ $\newline$ This gives us the initial condition for the recursive formula.
Express $f(n)$ in terms of $f(n-1)$ : Now, let's find $f(n)$ in terms of $f(n-1)$ . We know that: $\newline$ $f(n) = 45 \times \left(\frac{4}{5}\right)^{n-1}$ $\newline$ $f(n-1) = 45 \times \left(\frac{4}{5}\right)^{(n-1)-1} = 45 \times \left(\frac{4}{5}\right)^{n-2}$
Divide $f(n)$ by $f(n-1)$ : To express $f(n)$ in terms of $f(n-1)$ , we divide $f(n)$ by $f(n-1)$ :
$\frac{f(n)}{f(n-1)} = \frac{45 \times (\frac{4}{5})^{n-1}}{45 \times (\frac{4}{5})^{n-2}}$
Simplifying the right side, we get:
$\frac{f(n)}{f(n-1)} = (\frac{4}{5})$
Multiplying to get Recursive Formula: Multiplying both sides by $f(n-1)$ , we get the recursive formula: $\newline$ $f(n) = f(n-1) \times \left(\frac{4}{5}\right)$