Full solution

g(n)=80 \cdot\left(\frac{3}{4}\right)^{n}

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Complete the recursive formula of

g(n)

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g(1)= \square

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g(n)=g(n-1) \cdot \square

Given explicit formula: We are given the explicit formula for the sequence: $\newline$ $g(n) = 80 \times \left(\frac{3}{4}\right)^n$ $\newline$ To find the recursive formula, we need to express $g(n)$ in terms of $g(n-1)$ .
Find $g(1)$ : First, let's find $g(1)$ by substituting $n = 1$ into the explicit formula: $\newline$ $g(1) = 80 \times (\frac{3}{4})^1 = 80 \times \frac{3}{4} = 60$ $\newline$ So, $g(1) = 60$ .
Express in terms of $g(n-1)$ : Now, let's find $g(n)$ in terms of $g(n-1)$ . We know that: $\newline$ $g(n) = 80 \times \left(\frac{3}{4}\right)^n$ $\newline$ $g(n-1) = 80 \times \left(\frac{3}{4}\right)^{n-1}$ $\newline$ To express $g(n)$ in terms of $g(n-1)$ , we divide $g(n)$ by $g(n-1)$ : $\newline$ $\frac{g(n)}{g(n-1)} = \frac{80 \times \left(\frac{3}{4}\right)^n}{80 \times \left(\frac{3}{4}\right)^{n-1}}$
Simplify the equation: Simplify the right side of the equation: $\newline$ $g(n) / g(n-1) = (3/4)^n / (3/4)^{n-1}$ $\newline$ Using the property of exponents, $(a^m) / (a^n) = a^{m-n}$ , we get: $\newline$ $g(n) / g(n-1) = (3/4)^{n - (n-1)} = (3/4)^1 = 3/4$
Write recursive formula: Now we can write $g(n)$ in terms of $g(n-1)$ : $g(n) = g(n-1) \times \left(\frac{3}{4}\right)$ This is the recursive formula for the sequence.