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Math Help - Quadratic in Recurrence relation

  1. #1
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    Quadratic in Recurrence relation

    Hi, first time poster.

    Q. In the sequence u1, u2, u3,,un,
    u1 = 0, u2 = 3, u3 = 12 and un = a + bn + cn2
    Find the values of a, b and c.

    Attempt:
    If n = 1 then u1 = a + b(1) + c(1)2 = 0
    = a + b + c = 0
    If n = 2 then u2 = a + b(2) + c(2)2 = 3
    = a + 2b + 4c = 3
    If n = 3 then u3 = a + b(3) + c(3)2 = 12
    = a + 3b + 9c = 12

    = a + b + c = 0
    = a + 2b + 4c = 3
    = a + 3b + 9c = 12

    It's been a number of years since I've last studied maths. Can anyone help me on what comes next? Many thanks.
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  2. #2
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    Re: Quadratic in Recurrence relation

    a + b + c = 0
    a + 2b + 4c = 3
    a + 3b + 9c = 12
    subtract 1st equation from the 2nd ...

    b + 3c = 3

    subtract 2nd equation from the 3rd ...

    b + 5c = 9

    solve for b and c using these two equations
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Quadratic in Recurrence relation

    Are you able to use the linear combination method to solve systems?

    There's also one mistake, the last equation has to be: a+3b+6c=12
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  4. #4
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    Re: Quadratic in Recurrence relation

    Quote Originally Posted by GrigOrig99 View Post
    Hi, first time poster.

    Q. In the sequence u1, u2, u3,,un,
    u1 = 0, u2 = 3, u3 = 12 and un = a + bn + cn^2
    Find the values of a, b and c.

    Attempt:
    If n = 1 then u1 = a + b(1) + c(1)2 = 0
    = a + b + c = 0
    If n = 2 then u2 = a + b(2) + c(2)2 = 3
    = a + 2b + 4c = 3
    If n = 3 then u3 = a + b(3) + c(3)2 = 12
    = a + 3b + 9c = 12

    = a + b + c = 0
    = a + 2b + 4c = 3
    = a + 3b + 9c = 12

    It's been a number of years since I've last studied maths. Can anyone help me on what comes next? Many thanks.
    a+b+c=0 gives a+2b+4c=(a+b+c)+b+3c=3=0+3
    a+2b+4c=3 gives a+3b+9c=(a+2b+4c)+b+5c=12=3+9

    Then b+3c=3 gives b+5c=(b+3c)+2c=9=3+6
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  5. #5
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    Re: Quadratic in Recurrence relation

    Quote Originally Posted by Siron View Post
    Are you able to use the linear combination method to solve systems?

    There's also one mistake, the last equation has to be: a+3b+6c=12
    the OP meant a + bn + cn^2 ... the original system is correct.
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  6. #6
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    Re: Quadratic in Recurrence relation

    Fantastic! I've just managed to work it out. Thanks for all the help guys. Awesome site.
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