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Math Help - Swamped by a junior series (-2, 5, 24, 61, 122 ... ) formula?

  1. #1
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    Disregard

    Can anyone give me a formula for the series: -2, 5, 24, 61, 122 ....
    This is more of an association problem which shows my brains are fried. I'm a "has been" in the field and this posting is a total failure of my part. It is a pre Calculus 11th (junior level) grade. Got it Npower 3 -3
    Last edited by DanP; May 2nd 2012 at 07:39 PM.
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  2. #2
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    Re: Disregard

    nth term is n^3-3
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  3. #3
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    Re: Swamped by a junior series (-2, 5, 24, 61, 122 ... ) formula?

    For those of us who aren't as sharp as biffboy, here is an "algorithmic" approach- write these as a_0= -2, a_1= 5, a_2= 24, a_3= 61, a_4= 122, the "first differences" are \Delta_0= 5- (-2)= 7, \Delta_1= 24- 5= 19, \Delta_2= 61- 24= 37, and \Delta_3= 122- 61= 61; the "second differences" are \Delta^2_0= 19- 7= 12, \Delta^2_1= 37-19= 18, and \Delta^2_1= 61- 37= 24; the "third differences" are \Delta^3_0= 18- 12= 6, \Delta^3_1= 6, and \Delta^2_2= 6. Since the third differences are all the same, all succeeding differences are 0 and we can write this as a cubic equation.

    "Newton's divided difference formula" is similar to the Taylor's series but with finite differences rather than derivatives: f(0)+ \Delta f(0) n+ (\Delta^2 f(0)/2!)n(n-1)+ (\Delta^3 f(0)/3!)n(n-1)(n-2)+ \cdot\cdot\cdot.

    Here, that gives -2+ 7n+ (12/2)n(n-1)+ (6/6)n(n-1)(n-2)= -2+ 7n+ 6n^2- 6n+ n^3- 3n^2+ 2n
    n^3+ 3n^2+ 3n- 2= (n+1)^3- 3.

    There I started the indexing with n=0. Had I, instead, started with m= 1, so that n+1= m, that would be m^3- 3.
    Last edited by HallsofIvy; May 3rd 2012 at 08:54 AM.
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    Re: Swamped by a junior series (-2, 5, 24, 61, 122 ... ) formula?

    Great help!
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