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Math Help - nth term

  1. #1
    Member roshanhero's Avatar
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    nth term

    How to find nth term of the sequence 6+13+24+39+.....
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  2. #2
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    Re: nth term

    Hey roshanhero.

    For your problem without any more clarification there will be an infinite number of sequences that fit your description.

    You could find one by fitting a polynomial through them or use a program to generate a sequence.

    Without any more constraints, a unique sequence is not really possible.
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  3. #3
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    Re: nth term

    Your numbers seem random, there has to be a pattern if you want an expression for the nth term.
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  4. #4
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    Re: nth term

    Notice that the difference between the successive terms is 7,11,15 etc and that is an AP
    the sequence can be written as
    6 + ( 6 + 7) + ( 6+7+11) + ( 6+7+11+15 ) and so on now i am sure you can get the nth term
    Thanks from Shakarri
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  5. #5
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    Re: nth term

    I would do this using "Newton's divided difference formula". The numbers are a_0= 6, a_1= 13, a_2= 24, and a_3= 39. The "First Differences" are \Delta a_0= 13- 6= 7, \Delta a_1= 24- 13= 11, \Delta a_2= 39- 24= 15 and then the "Second Differences" are \Delta^2 a_0= 11- 7= 4, \Delta^2 a_1= 15- 11= 4. If we assume that the sequence continues so that the second differences are alway four (and the third and higher differences are 0) we can write, in analogy with the Taylor's series, a_n= a_0+ \Delta a_0 n+ (\Delta^2 a_0/2)n(n-1)= 6+ 7n+ (4/2)n(n- 1)= 6+ 7n+ 2n^2- 2n= 6+ 5n+ 2n^2.

    You will note that when n= 0, a_0= 6+ 5(0)+ 2(0^2)= 6, when n= 1, a_1= 6+ 5(1)+ 2(1^2)= 13, when n= 2, a_2= 6+ 5(2)+ 2(2^2)= 24, and when n= 3, a_3= 6+ 5(3)+ 2(3^2)= 39 exactly as we want.

    You could also use the fact that, since the second differences are constant, the sequence is a quadratic function of n: a_n= an^2+ bn+ c. Then use a_0= a(0)+ b(0)+ c= c= 6,
    a_1= a(1^2)+ b(1)+ c= a+ b+ c= 13 and a_2= a(2^2)+ b(2)c= 4a+ 2b+ c= 24 to solve for a, b, and c.
    Last edited by HallsofIvy; May 22nd 2013 at 12:05 PM.
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