218k views
4 votes
Which recursively defined function has a first term equal to 15 and a common difference of 3?

1 Answer

5 votes

Answer:

The recursively defined function is:


\left \{ {{f(1)\ =\ 15} \atop {f(n)=f(n-1)+3}} \right.

Explanation:

A recursively defined function has two parts:

  • The smallest argument, usually f (0) or f (1).
  • The nth argument, f (n) given f (n - 1), f (n - 2), etc.


\left \{ {{f(0)\ \text{or}\ f(1)\ =\ a} \atop {f(n)\ =\ f(n-1)\ +\ d}} \right.

In this case it is provided that:

The first term, f (1) = 15

Common difference, d = 3

Then the recursively defined function in this case is:


\left \{ {{f(1)\ =\ 15} \atop {f(n)=f(n-1)+3}} \right.

User Offbeatmammal
by
5.3k points