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

{f(1)=1f(2)=2f(n)=f(n−2)+f(n−1)f(3)=□ \begin{array}{l}\left\{\begin{array}{l}f(1)=1 \\ f(2)=2 \\ f(n)=f(n-2)+f(n-1)\end{array}\right. \\ f(3)=\square\end{array}

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Q. {f(1)=1f(2)=2f(n)=f(n−2)+f(n−1)f(3)=□ \begin{array}{l}\left\{\begin{array}{l}f(1)=1 \\ f(2)=2 \\ f(n)=f(n-2)+f(n-1)\end{array}\right. \\ f(3)=\square\end{array}
  1. Given initial conditions: We are given the initial conditions:\newlinef(1)=1f(1) = 1\newlinef(2)=2f(2) = 2\newlineAnd the recursive formula:\newlinef(n)=f(n−2)+f(n−1)f(n) = f(n-2) + f(n-1)\newlineWe need to find the value of f(3)f(3).
  2. Recursive formula: Using the recursive formula, we can calculate f(3)f(3) as follows:\newlinef(3)=f(3−2)+f(3−1)f(3) = f(3-2) + f(3-1)\newlinef(3)=f(1)+f(2)f(3) = f(1) + f(2)\newlinef(3)=1+2f(3) = 1 + 2\newlinef(3)=3f(3) = 3

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