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{[f(1)=4],[f(n)=f(n-1)*(-0.5)]:} Find an explicit formula for f(n). f(n)=

{f(1)=4,f(n)=f(n−1)⋅(−0.5)\begin{cases} f(1)=4, f(n)=f(n-1)\cdot(-0.5) \end{cases} Find an explicit formula for f(n)f(n).f(n)=f(n)=

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Q. {f(1)=4,f(n)=f(n−1)⋅(−0.5)\begin{cases} f(1)=4, f(n)=f(n-1)\cdot(-0.5) \end{cases} Find an explicit formula for f(n)f(n).f(n)=f(n)=
  1. Given Recursive Formula: Given the recursive formula for a sequence: f(1)=4f(1) = 4 and f(n)=f(n−1)×(−0.5)f(n) = f(n-1) \times (-0.5) for n > 1. We need to find an explicit formula for f(n)f(n).
  2. Observing Pattern of Sequence: To find an explicit formula, let's observe the pattern of the sequence for the first few terms. Starting with f(1)=4f(1) = 4:\newline- f(2)=f(1)×(−0.5)=4×(−0.5)=−2f(2) = f(1) \times (-0.5) = 4 \times (-0.5) = -2\newline- f(3)=f(2)×(−0.5)=−2×(−0.5)=1f(3) = f(2) \times (-0.5) = -2 \times (-0.5) = 1\newline- f(4)=f(3)×(−0.5)=1×(−0.5)=−0.5f(4) = f(3) \times (-0.5) = 1 \times (-0.5) = -0.5
  3. Identifying Exponential Pattern: Notice the pattern in the sequence: 4,−2,1,−0.5,ext...4, -2, 1, -0.5, ext{...} Each term is obtained by multiplying the previous term by −0.5-0.5. This suggests an exponential pattern with a base of −0.5-0.5 and a starting value of 44 when n=1n = 1.
  4. General Form of Exponential Function: The general form for an exponential function is f(n)=a⋅b(n−1)f(n) = a \cdot b^{(n - 1)}, where aa is the initial value, bb is the base of the exponential, and nn is the term number. In our case, a=4a = 4 (the first term) and b=−0.5b = -0.5 (the factor by which each term is multiplied to get the next term).
  5. Substituting Values into Formula: Substituting the values of aa and bb into the general form, we get the explicit formula for the sequence as f(n)=4×(−0.5)(n−1)f(n) = 4 \times (-0.5)^{(n - 1)}.

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