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Math Help - I don't understand sequences...

  1. #1
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    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
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    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...
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    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...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    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
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  5. #5
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    We're supposed to use n as 1,2,3,4,5,6 etc, no zero at the beggining.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    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
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  7. #7
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    What about the other one?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    ....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
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  9. #9
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    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
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    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,...
    Last edited by Jhevon; March 27th 2007 at 07:08 PM.
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  11. #11
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    Thanks a lot man, that really helped
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