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Match each explicit formula to its corresponding recursive formula.

[tex]\[
\begin{array}{ll}
f(n)=8(2)^{n-1} & f(n)=3+8(n-1) \\
f(n)=5(4)^{n-1} & f(n)=8+8(n-1) \\
f(n)=8(5)^{n-1} & \\
f(1)=8, f(n)=f(n-1)+8 \text{ for } n \geq 2 & \\
f(1)=8, f(n)=3+f(n-1) \text{ for } n \geq 2 & \\
f(1)=8, f(n)=5f(n-1) \text{ for } n \geq 2 &
\end{array}
\][/tex]


Sagot :

To match each explicit formula to its corresponding recursive formula, we will identify how each formula behaves and matches the recursive definitions.

1. Explicit formula: [tex]\( f(n) = 8(2)^{n-1} \)[/tex]
- Recursive formula: [tex]\( f(1) = 8, f(n) = 2 f(n-1) \text{ for } n \geq 2 \)[/tex]

2. Explicit formula: [tex]\( f(n) = 5(4)^{n-1} \)[/tex]
- Recursive formula: [tex]\( f(1) = 5, f(n) = 4 f(n-1) \text{ for } n \geq 2 \)[/tex]

3. Explicit formula: [tex]\( f(n) = 8 + 8(n-1) \)[/tex]
- Recursive formula: [tex]\( f(1) = 8, f(n) = f(n-1) + 8 \text{ for } n \geq 2 \)[/tex]

4. Explicit formula: [tex]\( f(n) = 3 + 8(n-1) \)[/tex]
- Recursive formula: [tex]\( f(1) = 8, f(n) = 3 + 8 f(n-1) \text{ for } n \geq 2 \)[/tex]

Now, let's summarize the pairs:
- [tex]\( f(n) = 8(2)^{n-1} \)[/tex] matches with [tex]\( f(1) = 8, f(n) = 2 f(n-1) \text { for } n \geq 2 \)[/tex]
- [tex]\( f(n) = 5(4)^{n-1} \)[/tex] matches with [tex]\( f(1) = 5, f(n) = 4 f(n-1) \text { for } n \geq 2 \)[/tex]
- [tex]\( f(n) = 8 + 8(n-1) \)[/tex] matches with [tex]\( f(1) = 8, f(n) = f(n-1) + 8 \text { for } n \geq 2 \)[/tex]
- [tex]\( f(n) = 3 + 8(n-1) \)[/tex] matches with [tex]\( f(1) = 8, f(n) = 3 + 8 f(n-1) \text { for } n \geq 2 \)[/tex]

Therefore, the matches are:
- [tex]\( 8(2)^{n-1} \)[/tex] and [tex]\( f(1)=8, f(n)=2f(n-1) \text{ for } n \geq 2 \)[/tex]
- [tex]\( 5(4)^{n-1} \)[/tex] and [tex]\( f(1)=5, f(n)=4f(n-1) \text{ for } n \geq 2 \)[/tex]
- [tex]\( 8 + 8(n-1) \)[/tex] and [tex]\( f(1)=8, f(n)=f(n-1)+8 \text{ for } n \geq 2 \)[/tex]
- [tex]\( 3 + 8(n-1) \)[/tex] and [tex]\( f(1)=8, f(n)=3+8 f(n-1) \text{ for } n \geq 2 \)[/tex]
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