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

Given Recursive Formula: Given the recursive formula for a sequence: f(1) = 4 and f(n) = f(n-1) * (-0.5) for n > 1. We need to find an explicit formula for f(n).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) = 4: - f(2) = f(1) * (-0.5) = 4 * (-0.5) = -2 - f(3) = f(2) * (-0.5) = -2 * (-0.5) = 1 - f(4) = f(3) * (-0.5) = 1 * (-0.5) = -0.5Identifying Exponential Pattern: Notice the pattern in the sequence: 4, -2, 1, -0.5, ... Each term is obtained by multiplying the previous term by -0.5. This suggests an exponential pattern with a base of -0.5 and a starting value of 4 when n = 1.General Form of Exponential Function: The general form for an exponential function is f(n) = a * b^(n - 1), where a is the initial value, b is the base of the exponential, and n is the term number. In our case, a = 4 (the first term) and b = -0.5 (the factor by which each term is multiplied to get the next term).Substituting Values into Formula: Substituting the values of a and b into the general form, we get the explicit formula for the sequence as f(n) = 4 * (-0.5)^(n - 1).

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

f(n)=

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

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Math Problems

Algebra 1

Write variable expressions for geometric sequences

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\begin{cases} f(1)=4, f(n)=f(n-1)\cdot(-0.5) \end{cases}

Find an explicit formula for

f(n)

f(n)=

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