Question

What would the first 4 terms be if the rule was f(n)=f(n-1)+8.f(0) = 0

Answer 1

From the problem, we have :

$f(n)=f(n-1)+8$

When substituting n = 1

$\begin{gathered} f(1)=f(1-1)+8 \\ f(1)=f(0)+8 \end{gathered}$

Substitute f(0) = 0

$f(1)=0+8$

Substitute n = 2

$\begin{gathered} f(2)=f(2-1)+8 \\ f(2)=f(1)+8 \\ f(2)=8+8 \\ f(2)=16 \end{gathered}$

Substitute n = 3

$\begin{gathered} f(3)=f(3-1)+8 \\ f(3)=f(2)+8 \\ f(3)=16+8 \\ f(3)=24 \end{gathered}$

Substitute n =4

$\begin{gathered} f(4)=f(4-1)+8 \\ f(4)=f(3)+8 \\ f(4)=24+8 \\ f(4)=32 \end{gathered}$

The first 4 terms are :

8, 16, 24, and 32