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$g(n)=-72*((1)/(6))^(n-1) Complete the recursive formula of g(n). {:[g(1)=◻],[g(n)=g(n-1).]:}$

$g(n)=-72 \cdot\left(\frac{1}{6}\right)^{n-1}$ $\newline$ Complete the recursive formula of $g(n)$ . $\newline$ $\begin{array}{l} g(1)=\square \\ g(n)=g(n-1) \cdot \square \end{array}$

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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}

\newline

Complete the recursive formula of

g(n)

\newline

\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}