Full solution

h(n)=63 \cdot\left(-\frac{1}{3}\right)^{n}

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Complete the recursive formula of

h(n)

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\begin{array}{l} h(1)=\square \\ h(n)=h(n-1) \cdot \square \end{array}

Given Explicit Formula: We are given the explicit formula for the sequence $h(n)$ as $h(n)=63\cdot\left(-\frac{1}{3}\right)^n$ . To find the recursive formula, we need to express $h(n)$ in terms of $h(n-1)$ .
Find $h(1)$ : First, let's find $h(1)$ by substituting $n=1$ into the explicit formula: $\newline$ $h(1) = 63 \times \left(-\frac{1}{3}\right)^{1} = 63 \times \left(-\frac{1}{3}\right) = -21.$ $\newline$ This gives us the initial condition for the recursive formula.
Express $h(n)$ in terms of $h(n-1)$ : Now, let's find $h(n)$ in terms of $h(n-1)$ . We know that $h(n) = 63 \times \left(-\frac{1}{3}\right)^n$ . We can express $h(n-1)$ similarly: $\newline$ $h(n-1) = 63 \times \left(-\frac{1}{3}\right)^{n-1}$ .
Divide $h(n)$ by $h(n-1)$ : To find the relationship between $h(n)$ and $h(n-1)$ , we can divide $h(n)$ by $h(n-1)$ : $\frac{h(n)}{h(n-1)} = \frac{63 \cdot \left(-\frac{1}{3}\right)^n}{63 \cdot \left(-\frac{1}{3}\right)^{n-1}}.$
Simplify the Equation: Simplify the right side of the equation by canceling out common factors and simplifying the powers of $-\frac{1}{3}$ : $\frac{h(n)}{h(n-1)} = \left(-\frac{1}{3}\right)^n / \left(-\frac{1}{3}\right)^{n-1} = -\frac{1}{3}.$
Express $h(n)$ in terms of $h(n-1)$ : Now we can express $h(n)$ in terms of $h(n-1)$ : $h(n) = h(n-1) \times -\left(\frac{1}{3}\right).$ This is the recursive formula for the sequence.