1. The initial condition is \(f(0) = 3\text{.}\) To get \(f(n+1)\) we would double the number of snails in the terrarium the previous year, which is given by \(f(n)\text{.}\) Thus \(f(n+1) = 2f(n)\text{.}\) The full recursive definition contains both of these, and would be written,

    \begin{equation*} f(0) = 3;~ f(n+1) = 2f(n)\text{.} \end{equation*}

  2. We are told that on day 0 you can do 7 push-ups, so \(g(0) = 7\text{.}\) The number of push-ups you can do on day \(n+1\) is 2 more than the number you can do on day \(n\text{,}\) which is given by \(g(n)\text{.}\) Thus

    \begin{equation*} g(0) = 7;~ g(n+1) = g(n) + 2\text{.} \end{equation*}

  3. Here \(h(0) = 1\text{.}\) To get the recurrence relation, think about how you can get \(h(n+1) = (n+1)!\) from \(h(n) = n!\text{.}\) If you write out both of these as products, you see that \((n+1)!\) is just like \(n!\) except you have one more term in the product, an extra \(n+1\text{.}\) So we have,

    \begin{equation*} h(0) = 1;~ h(n+1) = (n+1)\cdot h(n)\text{.} \end{equation*}