Answer:
[tex]f(n) = f(n-1) + 8[/tex] for [tex]n > 1[/tex]
Step-by-step explanation:
Given
[tex]f(1) = 3[/tex] -- First Term
[tex]\frac{1}{2}d = 4[/tex] --- half common difference
Required
Find the recursive rule
First, we calculate the common difference
[tex]\frac{1}{2}d = 4[/tex]
Multiply through by 2
[tex]2 * \frac{1}{2}d = 2 * 4[/tex]
[tex]d = 8[/tex]
The second term of the sequence is:
[tex]f(2) = 3 + 8 = 11[/tex]
The third term is:
[tex]f(3) = 11 + 8 = 20[/tex]
So, we have:
[tex]f(1) = 3[/tex]
[tex]f(2) = 3 + 8[/tex]
Substitute f(1) for 3
[tex]f(2) = f(1) + 8[/tex]
Express 1 as 2 - 1
[tex]f(2) = f(2-1) + 8[/tex]
Substitute n for 2
[tex]f(n) = f(n-1) + 8[/tex]
Similarly:
[tex]f(3) = 11 + 8[/tex]
Substitute f(2) for 11
[tex]f(3) = f(2) + 8[/tex]
Express 2 as 3 - 1
[tex]f(3) = f(3-1) + 8[/tex]
Substitute n for 3
[tex]f(n) = f(n-1) + 8[/tex]
Hence, the recursive is:
[tex]f(n) = f(n-1) + 8[/tex] for [tex]n > 1[/tex]
