# Coming up with recursive and closed form expressions

• KevinL
In summary, for the first sequence, the closed form expression is Cn = (1/4) * (1/3)^n-1 and the recursive formula is Cn = Cn-1/3. For the second sequence, the closed form expression is Dn = n! and the recursive formula is Dn = n*Dn-1. For the third sequence, the closed form expression is (3n-2)/(2n+1) and for the fourth sequence, the recursive formula for Qn is Qn = Qn-1 + Qn-2.

## Homework Statement

I am having some trouble coming up with recursive and closed form expressions of different sequences. I realize helping me with this would pretty much just be giving me the answer, but if anyone could also help me with how to think of the answers that would be nice.

1) Cn = (1/4, 1/12, 1/36, 1/108)

CF: ?
R: C(n-1)/3

2) Dn = (1, 2, 6, 24, 120)

CF: ?
R: n*D(n-1)

3) I only need the closed form for this. (1/3, 4/5, 7/7, 10/9, 13/11)

CF: ?/2n+1

4) Let (1, 1, 2, 3, 5, 8) be the Fibonacci sequence. Define a new sequence by Qn = F(n+1)/Fn

a. List the first several terms of Qn
(1, 2, 3/2, 5/3, 8/5)

b. Find a recusive formula for Qn

?

Let's start with the first one. You correctly noticed that the sequence is given by a recursion relation
$C_0 = 1/4, C_n = C_{n - 1} / 3$
(note how I wrote down explicitly what $C_{n-1}/3$ gives you by using an equality sign, and that I have included the first term which you need to calculate anything using the recursion relation).

Now make a few steps in your mind. If you want to calculate the second term in the series, you have to take the first one, 1/4, and divide it by 3. To go to the third, take the second, (1/4)/3 = 1/12, and divide it by 3. Now how can I go directly from the first term, 1/4, to the third one, 1/36?
Suppose that I want to go from the first one to the fifth...