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    Trying to revise for my maths exams(doing comp sci). Started on proof.

    I've been going over some revision questions and I have to find the method of differences to find Tn for the following:

    (a)2,6,12,20,30

    (b)7,22,45,76,115

    (c)3,20,63,144,275

    But I have no idea what the formula to use for this. Is it arithmetic progression or geometric progression? Can't find anything about this online or otherwise..
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    (Original post by theonn)
    Trying to revise for my maths exams(doing comp sci). Started on proof.

    I've been going over some revision questions and I have to find the method of differences to find Tn for the following:

    (a)2,6,12,20,30

    (b)7,22,45,76,115

    (c)3,20,63,144,275

    But I have no idea what the formula to use for this. Is it arithmetic progression or geometric progression? Can't find anything about this online or otherwise..
    For part a, can you find a relationship between the numbers? If so then try and find two series that denote it.
    Hint:
    Spoiler:
    Show
    Try factorising the number given e.g. 1x2 = 2, 2x3 = 6 Then find an algebraic relationship between these numbers
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    (Original post by joostan)
    For part a, can you find a relationship between the numbers? If so then try and find two series that denote it.
    Hint:
    Spoiler:
    Show
    Think about what it means to be an even number.
    The questions above are ones that were given to me in a class, a PHD student covers that class. He went over the first question and I got this:

    2-6-12-20-30
    -4-6-8-10--
    --2-2-2---

    The formula he gave us was Tn = a*n^b/b! +n
    a being 2, n being the term, b being the final row

    So for (1) it's Tn = 2n^2/2! + n

    I've tried to implement that formula into (2) but I'm really confused.
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    (Original post by theonn)
    The questions above are ones that were given to me in a class, a PHD student covers that class. He went over the first question and I got this:

    2-6-12-20-30
    -4-6-8-10--
    --2-2-2---

    The formula he gave us was Tn = a*n^b/b! +n
    a being 2, n being the term, b being the final row

    So for (1) it's Tn = 2n^2/2! + n

    I've tried to implement that formula into (2) but I'm really confused.
    Interesting, I found a different relationship.
    If you read my edited spoiler above and here, you'll see

    \ u_r = r(r-1) = r^2 - r \Rightarrow \ S_n = \displaystyle\sum_{r=1}^n r^2 - \displaystyle\sum_{r=1}^n r
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    (Original post by joostan)
    Interesting, I found a different relationship.
    If you read my edited spoiler above and here, you'll see

    \ u_r = r(r-1) = r^2 - r \Rightarrow \ S_n = \displaystyle\sum_{r=1}^n r^2 - \displaystyle\sum_{r=1}^n r
    So there isn't a concrete formula/way to work out Tn?
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    (Original post by theonn)
    So there isn't a concrete formula/way to work out Tn?
    There is. Having expressed this as two series you can then use the method of differences as you suggested in your title.
    There is a way to prove the method of differences/show that it's true or something using sigmas I think.
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    (Original post by joostan)
    There is. Having expressed this as two series you can then use the method of differences as you suggested in your title.
    There is a way to prove the method of differences/show that it's true or something using sigmas I think.
    Sorry what do you mean by two series? The formula doesn't work for (2) though?
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    (Original post by joostan)
    For part a, can you find a relationship between the numbers? If so then try and find two series that denote it.

    (Original post by joostan)
    There is. Having expressed this as two series you can then use the method of differences as you suggested in your title.
    The OP is clearly interested in finding the general term for the SEQUENCES, not their associated series.
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    (Original post by theonn)
    Trying to revise for my maths exams(doing comp sci). Started on proof.

    I've been going over some revision questions and I have to find the method of differences to find Tn for the following:

    (a)2,6,12,20,30

    (b)7,22,45,76,115

    (c)3,20,63,144,275

    But I have no idea what the formula to use for this. Is it arithmetic progression or geometric progression? Can't find anything about this online or otherwise..
    Firstly, ignore the other poster - he is talking about something else hence the confusion.

    With regard to your sequences - the point is that they follow neither arithmetic nor geometric progression. They follow a different type of progression hence the method your instructor gave you.

    This is how it works:

    You take your sequence and work out the first differences, then the second differences and so on. If you get to a point where the kth differences are the same (k is just some number of times you worked out differences) then the function T_n which gives the nth term of your sequence is a polynomial of degree k.

    So, for your example,

    2-6-12-20-30
    -4-6-8-10-- First difference
    --2-2-2--- Second difference
    The second differences are all the same so you know T_n is a polynomial of degree 2 (or quadratic) in the variable n

    i.e. T_n = an^2 + bn + c

    for some numbers a,b,c.

    At this stage, you can either substitute some numbers in from your sequence to work out a,b,c

    e.g. T_1 =2 is the first term of your sequence so 2 = T_1 = a + b + c
    T_2 = 6 so 6 = a(4) + b(2) + c

    and T_3 = 12 so 12 = a(9) + b(3) + c

    Therefore a,b,c are the unique solution to the linear system

    a + b + c = 2
    4a + 2b + c = 6
    9a + 3b + c = 12

    which is a=1, b=1,c=0.

    Therefore T_n = n^2 + n

    The thing about the formula you quoted was that it won't work in general... in general there may be a constant term (i.e. a non-zero value of c) however - the method I just outlined always works, so give it a go on the other examples.
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    (Original post by Mark85)
    i.e. T_n = an^2 + bn + c

    e.g. T_1 =2 is the first term of your sequence so 2 = T_1 = a + b + c
    T_2 = 6 so 6 = a(4) + b(2) + c

    and T_3 = 12 so 12 = a(9) + b(3) + c
    I've understood how you got T_1, not so sure about T_2 and T_3

    For T_2. You got the 4 from the first row of differences, then added 2 to make it 6?

    For T_3, I have no idea where you got the 9 from, or the 3
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    (Original post by theonn)
    I've understood how you got T_1, not so sure about T_2 and T_3

    For T_2. You got the 4 from the first row of differences, then added 2 to make it 6?

    For T_3, I have no idea where you got the 9 from, or the 3
    T_n is just the function that tells you the nth number in your original sequence - nothing to do with the differences.

    So for 2,6,12,20,30

    we have T_1 = 2, T_2 = 6, T_3 = 12

    just the first three numbers in the sequence. It is that simple.

    The only complicated thing was that I was assuming that in general T_n was a quadratic equation (quadratic because there was a common second difference; if there was only a common third difference it would be cubic etc.). It isn't hard to prove this but you could just take it on faith for the exercise if you prefer. I only know that T_n is quadratic because the sequence has second common differences - this won't be the case for any old sequence.

    So then, if I know that T_n = an^2 + bn + c for some numbers a,b,c

    which is just saying that T_n is quadratic in n, I can substitute in some values for T_n that I know to work out a,b and c.

    So, since T_n is the nth number of the sequence, I just substituted the first three numbers of the sequence i.e. I replaced n in

    T_n = an^2 + bn + c

    by 1,2 and 3 to get 3 equations from which I could then solve simultaneously to get a,b and c.
 
 
 
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