# Thread: Geometric Sequence, find a closed form.

1. ## Geometric Sequence, find a closed form.

Hi, I need help with this question:

Consider the following geometric sequence:

4, 6.8, 11.56, 19.652,...

Find a closed form for the sequence.

Could someone tell me what a closed form is and how to find it/work it out?

Thanks.

2. Take note that each one is 1.7 times the previous one.

A closed form is a general form which allows you to find each entry by plugging in, say, n.

$4\cdot (1.7)^{n}$

Now, plug in n=1, n=2, n=3, etc. and see what you get.

3. A "geometric sequence" is a sequence of numbers in which each term is "r" times the previous term. If "a" is the first term and you multiply every time by "r", you get a, ar, $ar^2$, $ar^3$, etc. A "closed form" for a sequence simply a formula in which you can plug in "n" and find the nth term: since a= $ar^0$ and ar= $ar^1$, if we start numbering the sequence at 0, each term is $ar^n$.

In your case, 4, 6.8, 11.56, 19.652,..., 6.8/4= 1.7, 11.56/6.8= 1.7, 19.652/11.56= 1.7 so this is, in fact, a geometric sequence- with "common ration, r= 1.7. The first term is a= 4. The closed form is, as galactus said, $a_n= 4(1.7)^n$.

$4(1.7)^0= 4$, $4(1.7)^1= 4(1.7)= 6.8$, $4(1.7)^2= 4(2.89)= 11.56$, $4(1.7)^3= 4(4.913)= 19.652$.

(Note: I said above "if we start numbering the sequence at 1". The formula, of course, depends not only on the numbers in the sequence but how each number is "labled". Some people prefer to start numbering at 1. If you do that each term would be $ar^{n-1}$.)