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Math Help - sequences formula

  1. #1
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    sequences formula

    http://keypress.com/documents/daa1/M...AA_MPYS_01.pdf

    WORK SHEET 1.3 Problem #4 a-h

    THANK YOU!
    Last edited by 13373l1t3; September 18th 2008 at 02:59 PM.
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  2. #2
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    Hello, 13373l1t3!

    They asked for the "recursive" formula.
    But if they want, say, the 21st term, we'd better have a "closed" form.


    4. Write a recursive formula to generate each sequence.
    Then find the indicated term.

    a)\;\;-15, -11, -7, -3, \hdots . Find the 10th term.

    The sequence increases by 4 at each stage.

    The recursive formula is: . a_n \:=\:a_{n-1} + 4
    (Each term is the preceding term plus 4.)

    The closed formula is: . a_n \:=\:-19 + 4n

    Therefore: . a_{10} \:=\:-19 + 4(10) \:=\:21



    b)\;\;1000, 100, 10, 1, \hdots . Find the 12th term.

    Each term is one-tenth of the preceding term.

    The formula is: . a_n \:=\:\frac{10,000}{10^n}

    Therefore: . a_{12} \:=\:\frac{10,000}{10^{12} }\:=\:10^{-8} \:=\:0.00000001



    c)\;\;17.25, 14.94, 12.63, 10.32, \hdots . Find the 15th term.

    The terms decrease by 2.31

    The formula is: . a_n \:=\:19.56 - 2.31n

    Therefore: . a_{15} \:=\:19.56 - 2.31(15) \:=\:-15.09



    d)\;\;0.3, -0.03, 0.003, -0.0003,\hdots . Find the 8th term.

    Each term is -0.1 times the preceding term.

    The formula is: . a_n \:=\:0.3(-0.1)^{n-1}

    Therefore: . a_8 \;=\;0.3(-0.1)^7 \:=\:0.00000003



    e)\;\;0, \frac{1}{6},\:\frac{1}{3},\:\frac{1}{2},\hdots . Find the 21st term.

    This is a sneaky one!
    Here is my thought process . . .

    The numerators seems to be all 1's.
    Then how come the first term has a numerator of 0 ?

    Then it hit me . . . the numerators are: . 0,1,2,3,4,\hdots
    . . and the denominators are all 6's !!

    Get it? .We have: . \frac{0}{6},\:\frac{1}{6},\:\frac{2}{6},\:\frac{3}  {6},\:\frac{4}{6},\hdots

    . . and that is why the sequence begins: . 0,\:\frac{1}{6},\:\frac{1}{3},\:\frac{1}{2},\:\hdo  ts

    The formula is: . a_n \;=\;\frac{n-1}{6}

    Therefore: . a_{21} \;=\;\frac{21-1}{6} \:=\:\frac{20}{6} \:=\:\frac{10}{3}



    f)\;\;-2, 4, -8, -16, \hdots . Find the 15th term.

    Each term is (-2) to a power.

    The formula is: . a_n \:=\:(-2)^n

    Therefore: . a_{15} \;=\;(-2)^{15} \;=\;-32,768

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