\begin{cases} f(1)=37 \\\\ f(n)=f(n-1)\cdot 0.3 \end{cases} Find an explicit formula for f(n).

Given initial condition: We are given the initial condition f(1) = 37 and the recursive formula f(n) = f(n-1) * 0.3. To find an explicit formula, we need to express f(n) in terms of n without the recursion.Identify pattern in terms: Let's look at the first few terms to identify a pattern: f(1) = 37 f(2) = f(1) * 0.3 = 37 * 0.3 f(3) = f(2) * 0.3 = (37 * 0.3) * 0.3 = 37 * 0.3^2 We can see that with each step, we multiply by an additional factor of 0.3.Generalize formula based on pattern: Based on the pattern, we can generalize that f(n) = 37 * 0.3^(n-1), where n is the term number.Verify formula for n=1: To verify this formula, we can check it for n=1: f(1) = 37 * 0.3^(1-1) = 37 * 0.3^0 = 37 * 1 = 37, which matches the given initial condition.Verify formula for n=2: Now, let's check it for another value, say n=2: f(2) = 37 * 0.3^(2-1) = 37 * 0.3^1 = 37 * 0.3 = 11.1, which matches the recursive definition f(2) = f(1) * 0.3.

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\begin{cases} f(1)=37 \\\\ f(n)=f(n-1)\cdot 0.3 \end{cases}

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Find an explicit formula for f(n).

Given initial condition: We are given the initial condition $f(1) = 37$ and the recursive formula $f(n) = f(n-1) \times 0.3$ . To find an explicit formula, we need to express $f(n)$ in terms of $n$ without the recursion.
Identify pattern in terms: Let's look at the first few terms to identify a pattern: $\newline$ $f(1) = 37$ $\newline$ $f(2) = f(1) \times 0.3 = 37 \times 0.3$ $\newline$ $f(3) = f(2) \times 0.3 = (37 \times 0.3) \times 0.3 = 37 \times 0.3^2$ $\newline$ We can see that with each step, we multiply by an additional factor of $0.3$ .
Generalize formula based on pattern: Based on the pattern, we can generalize that $f(n) = 37 \times 0.3^{(n-1)}$ , where $n$ is the term number.
Verify formula for $n=1$ : To verify this formula, we can check it for $n=1$ : $f(1) = 37 \times 0.3^{(1-1)} = 37 \times 0.3^0 = 37 \times 1 = 37$ , which matches the given initial condition.
Verify formula for $n=2$ : Now, let's check it for another value, say $n=2$ : $f(2) = 37 \times 0.3^{(2-1)} = 37 \times 0.3^1 = 37 \times 0.3 = 11.1$ , which matches the recursive definition $f(2) = f(1) \times 0.3$ .