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h(n)=-15*6^(n) Complete the recursive formula of h(n). {:[h(1)=◻],[h(n)=h(n-1).]:}

h(n)=−15⋅6n h(n)=-15 \cdot 6^{n} \newlineComplete the recursive formula of h(n) h(n) .\newlineh(1)=□h(n)=h(n−1)⋅□ \begin{array}{l} h(1)=\square \\ h(n)=h(n-1) \cdot \square \end{array}

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Q. h(n)=−15⋅6n h(n)=-15 \cdot 6^{n} \newlineComplete the recursive formula of h(n) h(n) .\newlineh(1)=□h(n)=h(n−1)⋅□ \begin{array}{l} h(1)=\square \\ h(n)=h(n-1) \cdot \square \end{array}
  1. Given Explicit Formula: We are given the explicit formula for the sequence h(n)=−15×6nh(n) = -15 \times 6^{n}. To find the recursive formula, we need to express h(n)h(n) in terms of h(n−1)h(n-1).
  2. Find h(1)h(1): First, let's find h(1)h(1) by substituting n=1n = 1 into the explicit formula:\newlineh(1)=−15×61=−15×6=−90.h(1) = -15 \times 6^{1} = -15 \times 6 = -90.\newlineThis gives us the initial condition for the recursive formula.
  3. Express in Terms of h(n−1-1): Now, let's find h(n)h(n) in terms of h(n−1)h(n-1). We know that h(n)=−15×6nh(n) = -15 \times 6^{n} and h(n−1)=−15×6n−1h(n-1) = -15 \times 6^{n-1}. To express h(n)h(n) in terms of h(n−1)h(n-1), we can divide h(n)h(n) by h(n−1)h(n-1):h(n)h(n−1)=−15×6n−15×6n−1=6n6n−1=6.\frac{h(n)}{h(n-1)} = \frac{-15 \times 6^{n}}{-15 \times 6^{n-1}} = \frac{6^{n}}{6^{n-1}} = 6.
  4. Recursive Formula: From the previous step, we have h(n)=6×h(n−1)h(n) = 6 \times h(n-1). This is the recursive formula for the sequence.

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