g(n)=80*((3)/(4))^(n) Complete the recursive formula of g(n). {:[g(1)=◻],[g(n)=g(n-1).]:}

Given explicit formula: We are given the explicit formula for the sequence: g(n) = 80 * (3/4)^n 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: g(1) = 80 * (3/4)^1 = 80 * 3/4 = 60 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: g(n) = 80 * (3/4)^n g(n-1) = 80 * (3/4)^(n-1) To express g(n) in terms of g(n-1), we divide g(n) by g(n-1): g(n) / g(n-1) = (80 * (3/4)^n) / (80 * (3/4)^(n-1))Simplify the equation: Simplify the right side of the equation: g(n) / g(n-1) = (3/4)^n / (3/4)^(n-1) Using the property of exponents, (a^m) / (a^n) = a^(m-n), we get: g(n) / g(n-1) = (3/4)^(n - (n-1)) = (3/4)^1 = 3/4Write recursive formula: Now we can write g(n) in terms of g(n-1): g(n) = g(n-1) * (3/4) This is the recursive formula for the sequence.

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Math Problems

Algebra 1

Write variable expressions for geometric sequences

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.