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

{g(1)=4g(n)=g(n1)(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}

Full solution

Q. {g(1)=4g(n)=g(n1)(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}
  1. Given initial condition and recursive formula: We are given the initial condition and the recursive formula for the sequence g(n)g(n):g(1)=4g(1) = 4g(n)=g(n1)(3)g(n) = g(n-1) \cdot (-3)To find g(3)g(3), we first need to find g(2)g(2).
  2. Calculating g(22): Using the recursive formula, we calculate g(22):\newlineg(22) = g(11) * 3-3\newlineg(22) = 44 * 3-3\newlineg(22) = 12-12
  3. Finding g(3)g(3) using g(2)g(2): Now that we have g(2)g(2), we can use it to find g(3)g(3):
    g(3)=g(2)(3)g(3) = g(2) \cdot (-3)
    g(3)=12(3)g(3) = -12 \cdot (-3)
    g(3)=36g(3) = 36

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