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{:[{[h(1)=9],[h(n)=h(n-1)*(1)/(9)]:}],[h(3)=◻]:}

{h(1)=9h(n)=h(n−1)⋅19h(3)=□ \begin{array}{l}\left\{\begin{array}{l}h(1)=9 \\ h(n)=h(n-1) \cdot \frac{1}{9}\end{array}\right. \\ h(3)=\square\end{array}

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Full solution

Q. {h(1)=9h(n)=h(n−1)⋅19h(3)=□ \begin{array}{l}\left\{\begin{array}{l}h(1)=9 \\ h(n)=h(n-1) \cdot \frac{1}{9}\end{array}\right. \\ h(3)=\square\end{array}
  1. Recursive formula and initial condition: The question prompt is asking for the value of h(3)h(3) given the recursive formula h(n)=h(n−1)×(19)h(n) = h(n-1) \times \left(\frac{1}{9}\right) and the initial condition h(1)=9h(1) = 9.
  2. Finding the value of h(2)h(2): First, we need to find the value of h(2)h(2) using the recursive formula and the initial condition.\newlineh(n)=h(n−1)×(19)h(n) = h(n-1) \times \left(\frac{1}{9}\right)\newlineh(2)=h(1)×(19)h(2) = h(1) \times \left(\frac{1}{9}\right)\newlineh(2)=9×(19)h(2) = 9 \times \left(\frac{1}{9}\right)\newlineh(2)=1h(2) = 1\newlineWe have found that h(2)=1h(2) = 1.
  3. Finding the value of h(33): Now, we can find the value of h(33) using the value of h(22) we just found.\newlineh(n) = h(n−1-1) ×\times 19\frac{1}{9}\newlineh(33) = h(22) ×\times 19\frac{1}{9}\newlineh(33) = 1×191 \times \frac{1}{9}\newlineh(33) = 19\frac{1}{9}\newlineWe have found that h(33) = 19\frac{1}{9}.

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