How do you find the recursive rule of the sequence 1,1,1/3, 1/4,1/15?

Answer 1

It is usually is a good idea to determine whether the sequence is an Arithmetic Sequence or a Geometric Sequence.

Arithmetic Sequence

In an Arithmetic Sequence the difference between one term and the next is a constant.

In other words, we just add the same value each time ... infinitely.

Geometric Sequences

In a Geometric Sequence each term is found by multiplying the previous term by a constant.

Looking at the given sequence, or computing the differences and the ratios, you can determine that it is neither.

This sequence has the very nice feature that all the numerators are 1. Use that information (often, you will have to convert fractions in order to compare them).

The denominators are:    1, 1, 3, 4, 15, ...

Putting the term numbers and the denominators together [note: I start with a₁; some people start with a₀]:

If you were to plot these numbers, you would see this unusual jump.

A recursive definition means that previous values of the sequence are used to determine the current value of the sequence. That is, to determine term six (that is, a₆) of the given sequence we need to use a previous term(s). Then, in order to determine that/those terms, we need to use still previous term(s) [note: this is called the “recursive case”]. When we get to the beginning, we use the “base case.”

For this sequence, in order to find term 6, we use the term number minus 1 and multiply by the value of the term before the preceding term (value of term 4 is 3, so 5*3 = 15). But we didn’t know the value of term 4, in a recursive definition we first had to find it as 3*1 [note that two terms must be defined as a base case]. The typical question might be, “Find the 10^th term in the sequence 1, 1, 3, 4, 15, …” 

The recursive definition is:

 a₁ = 1

 a₂ = 1

 a_n = n * a_n-2   for n > 2

To find a₁₀ = 10*a₈ = 10*(8*a₆) = 10*(8*(6*a₄) = 10*(8*(6*(4*a₂)

 a₂ = 1 is a base case

So, a₁₀ = 10*8*6*4*1 = 1920

The tenth term is 1920.

Now remember that we used denominators, so let's put that into the answer:

a₁ = 1/1

a₂ = 1/1

a_n = 1 / (n*a_n-2) for n > 2

Answer 2

This is a difficult problem so I suspect you typed something wrong.

Assuming you typed it correctly, I let my calculator do the work and found a polynomial with a 4th degree that will result in this sequence. But I doubt if that is what you are looking for. Please double check the typing.