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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⋅4n g(n)=-11 \cdot 4^{n} \newlineComplete the recursive formula of g(n) g(n) .\newlineg(1)=□ g(1)= \square \newlineg(n)=g(n−1)⋅□ g(n)=g(n-1) \cdot \square

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Q. g(n)=−11⋅4n g(n)=-11 \cdot 4^{n} \newlineComplete the recursive formula of g(n) g(n) .\newlineg(1)=□ g(1)= \square \newlineg(n)=g(n−1)⋅□ g(n)=g(n-1) \cdot \square
  1. Establish Base Case: To find the recursive formula for g(n)g(n), we need to express g(n)g(n) in terms of g(n−1)g(n-1). Let's start by finding the value of g(1)g(1) to establish the base case of the recursion.\newlineg(1)=−11×41g(1) = -11 \times 4^1\newlineg(1)=−11×4g(1) = -11 \times 4\newlineg(1)=−44g(1) = -44
  2. Express in Terms of Previous: Now, let's express g(n)g(n) in terms of g(n−1)g(n-1). We know that g(n)=−11×4ng(n) = -11 \times 4^n and g(n−1)=−11×4(n−1)g(n-1) = -11 \times 4^{(n-1)}. To find a relationship between g(n)g(n) and g(n−1)g(n-1), we can divide g(n)g(n) by g(n−1)g(n-1).
    g(n)g(n−1)=−11×4n−11×4(n−1)\frac{g(n)}{g(n-1)} = \frac{-11 \times 4^n}{-11 \times 4^{(n-1)}}
    Simplifying the right side, we get:
    g(n)g(n−1)=4n4(n−1)\frac{g(n)}{g(n-1)} = \frac{4^n}{4^{(n-1)}}
    g(n−1)g(n-1)00
    g(n−1)g(n-1)11
    g(n−1)g(n-1)22
    This means that g(n)g(n) is g(n−1)g(n-1)44 times g(n−1)g(n-1).
  3. Derive Recursive Formula: Now we can write the recursive formula for g(n)g(n) using the relationship we found:\newlineg(n)=4×g(n−1)g(n) = 4 \times g(n-1)\newlineHowever, we must not forget to include the negative sign from the original function. The correct recursive formula should be:\newlineg(n)=−4×g(n−1)g(n) = -4 \times g(n-1)
  4. Combine Base Case and Formula: Finally, we combine the base case and the recursive relationship to write the complete recursive formula for g(n)g(n):\begin{cases}g(1)=-44\g(n)=-4\cdot g(n-1)\end{cases}

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