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{:[{[g(1)=4],[g(n)=g(n-1)*(-3)]:}],[g(3)=◻]:}

$\begin{array}{l}\left\{\begin{array}{l}g(1)=4 \\ g(n)=g(n-1) \cdot(-3)\end{array}\right. \\ g(3)=\square\end{array}$

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Evaluate recursive formulas for sequences

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Q.

\begin{array}{l}\left\{\begin{array}{l}g(1)=4 \\ g(n)=g(n-1) \cdot(-3)\end{array}\right. \\ g(3)=\square\end{array}

Given initial condition and recursive formula: We are given the initial condition and the recursive formula for the sequence $g(n)$ : $g(1) = 4$ $g(n) = g(n-1) \cdot (-3)$ To find $g(3)$ , we first need to find $g(2)$ .
Calculating g( $2$ ): Using the recursive formula, we calculate g( $2$ ): $\newline$ g( $2$ ) = g( $1$ ) $*$ $-3$ $\newline$ g( $2$ ) = $4$ $*$ $-3$ $\newline$ g( $2$ ) = $-12$
Finding $g(3)$ using $g(2)$ : Now that we have $g(2)$ , we can use it to find $g(3)$ :
$g(3) = g(2) \cdot (-3)$
$g(3) = -12 \cdot (-3)$
$g(3) = 36$

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