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{[f(1)=-2],[f(2)=5],[f(n)=f(n-2)*f(n-1)]:} f(3)=

{f(1)=−2f(2)=5f(n)=f(n−2)⋅f(n−1)f(3)= \begin{array}{l}\left\{\begin{array}{l}f(1)=-2 \\ f(2)=5 \\ f(n)=f(n-2) \cdot f(n-1)\end{array}\right. \\ f(3)=\end{array}

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Q. {f(1)=−2f(2)=5f(n)=f(n−2)⋅f(n−1)f(3)= \begin{array}{l}\left\{\begin{array}{l}f(1)=-2 \\ f(2)=5 \\ f(n)=f(n-2) \cdot f(n-1)\end{array}\right. \\ f(3)=\end{array}
  1. Recursive formula: question_prompt: What is the value of f(3)f(3)?\newlineTo find f(3)f(3), we need to use the recursive formula given by f(n)=f(n−2)×f(n−1)f(n) = f(n-2) \times f(n-1). We have the values for f(1)f(1) and f(2)f(2), so we can calculate f(3)f(3) using these values.\newlinef(3)=f(1)×f(2)f(3) = f(1) \times f(2)\newlinef(3)=(−2)×5f(3) = (-2) \times 5\newlinef(3)=−10f(3) = -10

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