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{:[{[f(1)=15],[f(n)=f(n-1)*n]:}],[f(2)=◻]:}

{f(1)=15f(n)=f(n−1)⋅nf(2)=□ \begin{array}{l}\left\{\begin{array}{l}f(1)=15 \\ f(n)=f(n-1) \cdot n\end{array}\right. \\ f(2)=\square\end{array}

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