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{[g(1)=2.2],[g(n)=g(n-1)*(-5)]:} Find an explicit formula for g(n). g(n)=

{g(1)=2.2,g(n)=g(n−1)⋅(−5)\begin{cases} g(1)=2.2, g(n)=g(n-1)\cdot(-5) \end{cases} Find an explicit formula for g(n)g(n).g(n)=g(n)=

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Q. {g(1)=2.2,g(n)=g(n−1)⋅(−5)\begin{cases} g(1)=2.2, g(n)=g(n-1)\cdot(-5) \end{cases} Find an explicit formula for g(n)g(n).g(n)=g(n)=
  1. Initial Condition: The problem gives us the initial condition g(1)=2.2g(1) = 2.2 and a recursive formula g(n)=g(n−1)×(−5)g(n) = g(n-1) \times (-5). To find an explicit formula for g(n)g(n), we need to express g(n)g(n) in terms of nn without the need for previous terms.
  2. Apply Recursive Formula: Let's start by applying the recursive formula to the first few terms to see if we can identify a pattern.\newlineg(1)=2.2g(1) = 2.2\newlineg(2)=g(1)×(−5)=2.2×(−5)g(2) = g(1) \times (-5) = 2.2 \times (-5)\newlineg(3)=g(2)×(−5)=(2.2×(−5))×(−5)g(3) = g(2) \times (-5) = (2.2 \times (-5)) \times (-5)\newlineWe can see that each term is the previous term multiplied by −5-5.
  3. Express in Terms of n: Now let's express g(n)g(n) in terms of g(1)g(1):
    g(2)=g(1)×(−5)g(2) = g(1) \times (-5)
    g(3)=g(2)×(−5)=g(1)×(−5)2g(3) = g(2) \times (-5) = g(1) \times (-5)^2
    g(4)=g(3)×(−5)=g(1)×(−5)3g(4) = g(3) \times (-5) = g(1) \times (-5)^3
    ...
    g(n)=g(1)×(−5)(n−1)g(n) = g(1) \times (-5)^{(n-1)}
  4. Substitute Value into Formula: Since we know g(1)=2.2g(1) = 2.2, we can substitute this value into our formula: g(n)=2.2×(−5)(n−1)g(n) = 2.2 \times (-5)^{(n-1)} This is the explicit formula for g(n)g(n).

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