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$h(n) = -15\cdot 6^{n}$ $\newline$ Complete the recursive formula of $h(n)$ . $\newline$ $h(1)=\square$ $\newline$ $h(n)=h(n-1) \cdot \square$

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Transformations of functions

Full solution

Q.

h(n) = -15\cdot 6^{n}

\newline

Complete the recursive formula of

h(n)

.

\newline

h(1)=\square

\newline

h(n)=h(n-1) \cdot \square

Find h( $1$ ): To find the recursive formula, we need to express $h(n)$ in terms of $h(n-1)$ . Let's start by finding $h(1)$ . $\newline$ $h(n) = -15\cdot6^n$ $\newline$ $h(1) = -15\cdot6^1$ $\newline$ $h(1) = -15\cdot6$ $\newline$ $h(1) = -90$
Establish Relationship: Now we need to find the relationship between $h(n)$ and $h(n-1)$ . We know that $h(n) = -15\cdot6^n$ and $h(n-1) = -15\cdot6^{(n-1)}$ . To find the multiplier that gets us from $h(n-1)$ to $h(n)$ , we divide $h(n)$ by $h(n-1)$ .
$\frac{h(n)}{h(n-1)} = \frac{-15\cdot6^n}{-15\cdot6^{(n-1)}}$
$\frac{h(n)}{h(n-1)} = \frac{6^n}{6^{(n-1)}}$
$h(n-1)$ $0$
$h(n-1)$ $1$
$h(n-1)$ $2$
So, $h(n-1)$ $3$

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