Question

spaced practice
determine the first four terms of each sequence when n is a whole
number greater than 1.

1. ( f(n) = 5f(n - 1) ) and ( f(1) = 3 )
2. ( h(n) = \frac{1}{4}f(n - 1) ) and ( f(1) = 80 )

Explanation:

Problem 1: \( f(n) = 5f(n - 1) \) and \( f(1) = 3 \)

Step 1: Find \( f(1) \)

Given \( f(1) = 3 \).

Step 2: Find \( f(2) \)

Substitute \( n = 2 \) into \( f(n) = 5f(n - 1) \). So \( f(2) = 5f(2 - 1) = 5f(1) \). Since \( f(1) = 3 \), then \( f(2) = 5\times3 = 15 \).

Step 3: Find \( f(3) \)

Substitute \( n = 3 \) into \( f(n) = 5f(n - 1) \). So \( f(3) = 5f(3 - 1) = 5f(2) \). Since \( f(2) = 15 \), then \( f(3) = 5\times15 = 75 \).

Step 4: Find \( f(4) \)

Substitute \( n = 4 \) into \( f(n) = 5f(n - 1) \). So \( f(4) = 5f(4 - 1) = 5f(3) \). Since \( f(3) = 75 \), then \( f(4) = 5\times75 = 375 \).

Step 1: Find \( h(1) \)

Assuming \( h(1) = 80 \) (since the problem says \( f(1) = 80 \) but the function is \( h(n) \), probably a typo).

Step 2: Find \( h(2) \)

Substitute \( n = 2 \) into \( h(n) = \frac{1}{4}h(n - 1) \). So \( h(2) = \frac{1}{4}h(2 - 1) = \frac{1}{4}h(1) \). Since \( h(1) = 80 \), then \( h(2) = \frac{1}{4}\times80 = 20 \).

Step 3: Find \( h(3) \)

Substitute \( n = 3 \) into \( h(n) = \frac{1}{4}h(n - 1) \). So \( h(3) = \frac{1}{4}h(3 - 1) = \frac{1}{4}h(2) \). Since \( h(2) = 20 \), then \( h(3) = \frac{1}{4}\times20 = 5 \).

Step 4: Find \( h(4) \)

Substitute \( n = 4 \) into \( h(n) = \frac{1}{4}h(n - 1) \). So \( h(4) = \frac{1}{4}h(4 - 1) = \frac{1}{4}h(3) \). Since \( h(3) = 5 \), then \( h(4) = \frac{1}{4}\times5 = \frac{5}{4} = 1.25 \).

Answer:

The first four terms of the sequence \( f(n) \) are \( 3, 15, 75, 375 \).

Problem 2: \( h(n) = \frac{1}{4}f(n - 1) \) and \( f(1) = 80 \) (assuming \( h(1) \) is related to \( f(1) \), maybe a typo and should be \( h(1) = 80 \))