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$g(n)=-11*4^(n) Complete the recursive formula of g(n). {:[g(1)=◻],[g(n)=g(n-1).]:}$

$g(n)=-11 \cdot 4^{n}$ $\newline$ Complete the recursive formula of $g(n)$ . $\newline$ $g(1)= \square$ $\newline$ $g(n)=g(n-1) \cdot \square$

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Write variable expressions for geometric sequences

Full solution

Q.

g(n)=-11 \cdot 4^{n}

\newline

Complete the recursive formula of

g(n)

.

\newline

g(1)= \square

\newline

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

Establish Base Case: To find the recursive formula for $g(n)$ , we need to express $g(n)$ in terms of $g(n-1)$ . Let's start by finding the value of $g(1)$ to establish the base case of the recursion. $\newline$ $g(1) = -11 \times 4^1$ $\newline$ $g(1) = -11 \times 4$ $\newline$ $g(1) = -44$
Express in Terms of Previous: Now, let's express $g(n)$ in terms of $g(n-1)$ . We know that $g(n) = -11 \times 4^n$ and $g(n-1) = -11 \times 4^{(n-1)}$ . To find a relationship between $g(n)$ and $g(n-1)$ , we can divide $g(n)$ by $g(n-1)$ .
$\frac{g(n)}{g(n-1)} = \frac{-11 \times 4^n}{-11 \times 4^{(n-1)}}$
Simplifying the right side, we get:
$\frac{g(n)}{g(n-1)} = \frac{4^n}{4^{(n-1)}}$
$g(n-1)$ $0$
$g(n-1)$ $1$
$g(n-1)$ $2$
This means that $g(n)$ is $g(n-1)$ $4$ times $g(n-1)$ .
Derive Recursive Formula: Now we can write the recursive formula for $g(n)$ using the relationship we found: $\newline$ $g(n) = 4 \times g(n-1)$ $\newline$ However, we must not forget to include the negative sign from the original function. The correct recursive formula should be: $\newline$ $g(n) = -4 \times g(n-1)$
Combine Base Case and Formula: Finally, we combine the base case and the recursive relationship to write the complete recursive formula for $g(n)$ :\begin{cases}g(1)=-44\g(n)=-4\cdot g(n-1)\end{cases}

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