g(n)=-72*((1)/(6))^(n-1) 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) = -72 * (1/6)^(n - 1) To find the recursive formula, we need to express g(n) in terms of g(n-1). First, let's find g(1) by substituting n = 1 into the explicit formula. g(1) = -72 * (1/6)^(1 - 1) g(1) = -72 * (1/6)^0 g(1) = -72 * 1 g(1) = -72Find g(1): Now, let's find g(n) in terms of g(n-1). We know that g(n) = -72 * (1/6)^(n - 1). Let's write g(n-1) using the same formula: g(n-1) = -72 * (1/6)^((n-1) - 1) g(n-1) = -72 * (1/6)^(n - 2)Express g(n) in terms of g(n-1): To express g(n) in terms of g(n-1), we can divide g(n) by g(n-1): g(n) / g(n-1) = (-72 * (1/6)^(n - 1)) / (-72 * (1/6)^(n - 2)) Simplifying the right side, we get: g(n) / g(n-1) = (1/6)^(n - 1) / (1/6)^(n - 2) g(n) / g(n-1) = (1/6)Write Recursive Formula: Now, we can write g(n) in terms of g(n-1): g(n) = g(n-1) * (1/6)We have found g(1) and the relationship between g(n) and g(n-1). Therefore, the recursive formula for the sequence is: {:[g(1)=-72],[g(n)=g(n-1)*(1/6)]:}

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Math Problems

Algebra 1

Write variable expressions for geometric sequences

Full solution

g(n)=-72 \cdot\left(\frac{1}{6}\right)^{n-1}

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Complete the recursive formula of

g(n)

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\begin{array}{l} g(1)=\square \\ g(n)=g(n-1) \cdot \square \end{array}

Given Explicit Formula: We are given the explicit formula for the sequence: $\newline$ $g(n) = -72 \times \left(\frac{1}{6}\right)^{n - 1}$ $\newline$ To find the recursive formula, we need to express $g(n)$ in terms of $g(n-1)$ . $\newline$ First, let's find $g(1)$ by substituting $n = 1$ into the explicit formula. $\newline$ $g(1) = -72 \times \left(\frac{1}{6}\right)^{1 - 1}$ $\newline$ $g(1) = -72 \times \left(\frac{1}{6}\right)^0$ $\newline$ $g(1) = -72 \times 1$ $\newline$ $g(1) = -72$
Find $g(1)$ : Now, let's find $g(n)$ in terms of $g(n-1)$ . We know that $g(n) = -72 \times (\frac{1}{6})^{(n - 1)}$ . Let's write $g(n-1)$ using the same formula: $g(n-1) = -72 \times (\frac{1}{6})^{((n-1) - 1)}$ $g(n-1) = -72 \times (\frac{1}{6})^{(n - 2)}$
Express $g(n)$ in terms of $g(n-1)$ : To express $g(n)$ in terms of $g(n-1)$ , we can divide $g(n)$ by $g(n-1)$ : $\frac{g(n)}{g(n-1)} = \frac{-72 \times (\frac{1}{6})^{n - 1}}{-72 \times (\frac{1}{6})^{n - 2}}$ Simplifying the right side, we get: $\frac{g(n)}{g(n-1)} = \frac{(\frac{1}{6})^{n - 1}}{(\frac{1}{6})^{n - 2}}$ $\frac{g(n)}{g(n-1)} = \frac{1}{6}$
Write Recursive Formula: Now, we can write $g(n)$ in terms of $g(n-1)$ : $g(n) = g(n-1) \times \left(\frac{1}{6}\right)$
Write Recursive Formula: Now, we can write $g(n)$ in terms of $g(n-1)$ : $g(n) = g(n-1) \times \left(\frac{1}{6}\right)$ We have found $g(1)$ and the relationship between $g(n)$ and $g(n-1)$ . Therefore, the recursive formula for the sequence is:\begin{cases}g(1)=-72\g(n)=g(n-1)\times\left(\frac{1}{6}\right)\end{cases}