Question

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) = (1/4)f(n - 1) and f(1) = 80

Explanation:

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

Step 1: Find \( f(2) \)

We know \( f(1) = 3 \), and for \( n = 2 \), we use the formula \( f(n)=5f(n - 1) \). So \( f(2)=5\times f(1) \). Substituting \( f(1) = 3 \), we get \( f(2)=5\times3 = 15 \).

Step 2: Find \( f(3) \)

For \( n = 3 \), using the formula \( f(n)=5f(n - 1) \), we have \( f(3)=5\times f(2) \). Since \( f(2) = 15 \), then \( f(3)=5\times15=75 \).

Step 3: Find \( f(4) \)

For \( n = 4 \), using the formula \( f(n)=5f(n - 1) \), we have \( f(4)=5\times f(3) \). Since \( f(3) = 75 \), then \( f(4)=5\times75 = 375 \).

Step 1: Find \( h(2) \)

We know \( h(1) = 80 \), and for \( n = 2 \), we use the formula \( h(n)=\frac{1}{4}h(n - 1) \). So \( h(2)=\frac{1}{4}\times h(1) \). Substituting \( h(1) = 80 \), we get \( h(2)=\frac{1}{4}\times80 = 20 \).

Step 2: Find \( h(3) \)

For \( n = 3 \), using the formula \( h(n)=\frac{1}{4}h(n - 1) \), we have \( h(3)=\frac{1}{4}\times h(2) \). Since \( h(2) = 20 \), then \( h(3)=\frac{1}{4}\times20 = 5 \).

Step 3: Find \( h(4) \)

For \( n = 4 \), using the formula \( h(n)=\frac{1}{4}h(n - 1) \), we have \( h(4)=\frac{1}{4}\times h(3) \). 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 \( f(1)=3 \), \( f(2)=15 \), \( f(3)=75 \), \( f(4)=375 \).

Problem 2: \( h(n)=\frac{1}{4}f(n - 1) \) (assuming it's a typo and should be \( h(n)=\frac{1}{4}h(n - 1) \)) and \( h(1) = 80 \)