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{[f(1)=-1.8],[f(n)=f(n-1)*9]:}
Find an explicit formula for 
f(n).

f(n)=

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{f(1)=1.8f(n)=f(n1)9 \left\{\begin{array}{l} f(1)=-1.8 \\ f(n)=f(n-1) \cdot 9 \end{array}\right. \newlineFind an explicit formula for f(n) f(n) .\newlinef(n)= f(n)= \newline \square

Full solution

Q. {f(1)=1.8f(n)=f(n1)9 \left\{\begin{array}{l} f(1)=-1.8 \\ f(n)=f(n-1) \cdot 9 \end{array}\right. \newlineFind an explicit formula for f(n) f(n) .\newlinef(n)= f(n)= \newline \square
  1. Base Case: We know f(1)=1.8f(1) = -1.8. This is our base case.
  2. Recursive Formula: The recursive formula is f(n)=f(n1)×9f(n) = f(n-1) \times 9. This means each term is 99 times the previous term.
  3. Explicit Formula: To find an explicit formula, we need to express f(n)f(n) in terms of nn and the base case f(1)f(1).
  4. Pattern Analysis: Since f(n)=f(n1)×9f(n) = f(n-1) \times 9, we can write f(2)=f(1)×9f(2) = f(1) \times 9, f(3)=f(2)×9f(3) = f(2) \times 9, and so on.
  5. Final Formula: This pattern shows that f(n)=f(1)×9(n1)f(n) = f(1) \times 9^{(n-1)}.
  6. Substitution: Substitute f(1)=1.8f(1) = -1.8 into the formula: f(n)=1.8×9(n1)f(n) = -1.8 \times 9^{(n-1)}.

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