Sequence formula help?

Question

How would I find the formula for this sequence: 1, 3/2, 5/4, 7/8, 9/16, 11/32, 13/64, 15/128. I know the that 2 is added to the numerator, and the denominator is multiplied by 2. I don't know where to go from there.

Answer 1

Hmm, didn't you just answer your own question...?

The nth number is (2n - 1)/2^(n-1). That's is just converting your last statement to "math-ese".

Answer 2

Alright, let's evaluate this step by step:

First let's look at the numerators to develop a rule for them first:

a_1 = 1
a_2 = 3

How do we get 3:

1 + 2 = 3

a_3 = 5

3 + 2 = 5

So we know the rule is this:

n + 2

That's the numerator, now we need a rule for the denominator:

a_1 = 1
a_2 = 2

How do we get two, well there are a few ways:

1*2 = 2
2^1 = 2

a_3 = 4

Now let's try the methods again:

2*2 = 4
2^2 = 4

a_4 = 8:

4*2 = 8
2^3 = 8

Here is where the first method fails: 4*2 = 2n while the rest are 2(n - 1). So we can conclude that the denominator is:

2^(n - 1)

So we see that the definition for this sequence is:

(n + 2) / 2^(n - 1)

And there you go.

Answer 3

Let n = the number of the item in the sequence, and x be the result of the formula, so that when n=1, x=1; when n=2, x=3/2, etc.

So you're looking for a formula to which you can plug in the "n" and get the "x".

x = (2n-1) / 2^(n-1)
when n=1; x = (2*1 - 1) / 2^(1-1) = (2-1)/2^0 = 1/1 = 1
when n=2; x = (2*2 - 1) / 2^(2-1) = (4-1)/2^1 = 3/2
when n=3; x = (2*3 - 1) / 2^(3-1) = (6-1)/2^2 = 5/4
when n=11; x = (2*11 - 1)/2^(11-1) = (22-1)/2^10 = 21/1024

Answer 4

You are right in your observation the numerator sequence is the odd numbers and the denominator sequence is the powers of 2.
What you want is a formula for the n-th element in the sequence.

Let s(n) be the n-the element in the sequence.
Then s(1) = 1/1 = 1

s(2) = (1+2)/2
s(3) = (s(2)+2)/2^2
... etc...

so s(n) = (s(n-1) +2)/2^(n-1)

Note that s(n-1) + 2 = 1 + 2 (n-1) so plugging in,

s(n) = (2n - 1)/ (2^(n-1)) for n >= 1.

Answer 5

You would generalize: starting with n=1, the nth term will have 2(n-1)+1 in the numerator and
2^(n-1) in the denominator. So that would form your general term for the sequence, valid from n=1 to whatever.

Answer 6

in line with threat this series is lacking words... i assume it is... 0, a million/3, 2/4, 3/5, 4/6, 5/7, 6/8, 7/9, 8/10, 9/11, 10/12, 11/thirteen, 12/14, thirteen/15, 14/sixteen, 15/17... etc... Then the formula is (n-a million) /(n+a million) 0 is the 1st (a million-a million)/(a million+a million) = 0 3/5 is the 4th... (4-a million)/(4+a million) = 3/5 8/10 is the ninth... (9-a million)/(9=a million) = 8/10 15/17 is the sixteenth... (sixteen-a million)/(sixteen+a million) = 15/17 24/26 is the twenty 5th... (25-a million)/(25+a million) = 24/26 .... ok!