# I don't understand sequences...

• Mar 27th 2007, 03:27 PM
Trentt
I don't understand sequences...
Find the pattern and write formula
3, 6, 12, 24, 48, 96

I understand that the number doubles each time, but I am quite confused as to how to figure out the formula for this.

Another problem that I'm having trouble with:

32, -16, 8, -4, 2 ,-1

Thanks a lot
• Mar 27th 2007, 03:31 PM
Jhevon
Quote:

Originally Posted by Trentt
Find the pattern and write formula
3, 6, 12, 24, 48, 96

3, 6, 12, 24, 48, 96 ...
= 3*1, 3*2, 3*2*2, 3*2*2*2, ...
= 3*2^0, 3*2, 3*2^2, 3*2^3...

so our formula is

a_n = 3*2^n for n = 0,1,2,3,4...
• Mar 27th 2007, 03:33 PM
Jhevon
Quote:

Originally Posted by Trentt
32, -16, 8, -4, 2 ,-1

32, -16, 8, -4, 2 ,-1 ...
= 32*1, 32*(-1/2), 32*(-1/2)*(-1/2),...
= 32(-1/2)^0, 32(-1/2)^2, 32(-1/2)^3...

so a_n = 32(-1/2)^n for n = 0,1,2,3,4...
• Mar 27th 2007, 03:34 PM
Jhevon
Quote:

Originally Posted by Trentt
Find the pattern and write formula
3, 6, 12, 24, 48, 96

I understand that the number doubles each time, but I am quite confused as to how to figure out the formula for this.

Another problem that I'm having trouble with:

32, -16, 8, -4, 2 ,-1

Thanks a lot

so these are geometric series.

the first has the first term 3 and common ratio 2

the second has the first term 32 and common ration -1/2
• Mar 27th 2007, 04:01 PM
Trentt
We're supposed to use n as 1,2,3,4,5,6 etc, no zero at the beggining.
• Mar 27th 2007, 04:11 PM
Jhevon
Quote:

Originally Posted by Trentt
We're supposed to use n as 1,2,3,4,5,6 etc, no zero at the beggining.

ok, just change the power to n-1
• Mar 27th 2007, 04:14 PM
Trentt
• Mar 27th 2007, 04:20 PM
Jhevon
Quote:

Originally Posted by Trentt
....That doesnt even work

for the first:

a_n = 3*2^(n-1) for n = 1,2,3,4...

the terms are: 3, 6, 12, 24, ...
that looks like the first sequence to me

for the second:

a_n = 32(-1/2)^(n-1) for n = 1,2,3,4...

the terms are: 32, -16, 8, -4,...
that looks like the second sequence to me
• Mar 27th 2007, 04:32 PM
Trentt
I really don't understand how you find the pattern in the sequence, that is my main trouble. Perhaps you could explain to me how to find the pattern in:

1, 3/2, 5/4, 7/8, 9/16, 11/32
• Mar 27th 2007, 04:49 PM
Jhevon
Quote:

Originally Posted by Trentt
I really don't understand how you find the pattern in the sequence, that is my main trouble. Perhaps you could explain to me how to find the pattern in:

1, 3/2, 5/4, 7/8, 9/16, 11/32

well you found the pattern yourself in the first two, you realized one was being doubled and the other was being halved. so i tried to write the terms out in terms of what was happening. sorry you didn't get it. let's try this one step by step.

1, 3/2, 5/4, 7/8, 9/16, 11/32....

ok, i notice that the denominator keep increasing by a factor of 2, we can see this if we focus only on the denominator, say like this, call all the tops 1 for the time being

1*(1/2) = 1/2

1/2 * 1/2 = 1/4

1/4 * 1/2 = 1/8

notice that for the denominators, you get every new one by multiplying the old one by 2, that's the pattern for the denominators. so i expect my common ratio to look something like (a/2)^n

now focus on the numerators, how are they changing?

we have 1 then 3 then 5 then 7 then 9...notice a pattern? these are all consecutive odd numbers. odd numbers are given by the formula, 2n+1, so we expect our formula to have the form 2n+1(1/2), but now what are the powers? we want to use n=1,2,3,4,5...

so for the first i want 1*1 then 1*3/2 then 1*5/4

so if we say (2n+1)(1/2)^n, for n=1,2,3,4,5... we get the sequence

3/2, 5/4, 7/8,... almost our sequence, but not completely, we want the term before 3/2 to be included, so we make all the n's into n-1's.

so we get a_n = (2(n - 1)+1)(1/2)^(n-1), for n = 1,2,3,4,5...
or a little more nicey, a_n = (2n-1)(1/2)^(n-1) for n=1,2,3,4,5...
or maybe even (2n - 1)/2^(n-1)

the sequence is: 1, 3/2, 5/4,...
• Mar 27th 2007, 05:55 PM
Trentt
Thanks a lot man, that really helped :)