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Math Help - Order of Differences, Sequences, Can't find information about it

  1. #1
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    Thumbs up Order of Differences, Sequences, Can't find information about it

    Hi Forum
    I've found this Order of Differences problem, which I can't find information on.
    This may be recurrent on sequences, maybe I got the name of it wrong.
    Anyway

    We are to find the sum of the first 7 terms in this sequences.



    Code:
    1
        > 2
    3        > 1
        > 3
    6        > 1
        > 4
    10      > 1
        > 5
    15
    The solution says something about T(n) being of the second degree.
    Then later it states
    1=a+b+c
    3=4a+2b+c
    6=9a+3b+c

    The 1, 3 and 6 part is related to the number of points in the triangle but what does the other part means?
    This seems like a complex sequence at first.
    Can someone help?

    Thanks
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  2. #2
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    Re: Order of Differences, Sequences, Can't find information about it

    Quote Originally Posted by Zellator View Post
    Hi Forum
    I've found this Order of Differences problem, which I can't find information on.
    This may be recurrent on sequences, maybe I got the name of it wrong.
    Anyway

    We are to find the sum of the first 7 terms in this sequences.



    Code:
    1
        > 2
    3        > 1
        > 3
    6        > 1
        > 4
    10      > 1
        > 5
    15
    The solution says something about T(n) being of the second degree.
    Then later it states
    1=a+b+c
    3=4a+2b+c
    6=9a+3b+c

    The 1, 3 and 6 part is related to the number of points in the triangle but what does the other part means?
    This seems like a complex sequence at first.
    Can someone help?

    Thanks
    1. The numbers which are calculated by T(n) belong to an arithmetic sequence of order (or degree) 2 because the 2nd column of the differences belong to a constant sequence (which has the order 0).

    2. Therefore the general form of T(n) is
    T(n)=an^2+bn+c

    3. If you plug in n and T(n) into this genral equation you'll get the system of simultaneous equations in (a, b, c):
    \left|\begin{array}{rcl}a+b+c&=&1 \\4a+2b+c&=&3 \\9a+3b+c &=&6 \end{array}\right.
    Solve this system and you should come out with (a, b, c)=\left(\frac12, \frac12, 0\right)

    4. Plug in n = 7 and calculate T(7)
    Spoiler:
    T(7) = 28
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  3. #3
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    Re: Order of Differences, Sequences, Can't find information about it

    Quote Originally Posted by Zellator View Post
    Hi Forum
    I've found this Order of Differences problem, which I can't find information on.
    This may be recurrent on sequences, maybe I got the name of it wrong.
    Anyway

    We are to find the sum of the first 7 terms in this sequences.



    Code:
    1
        > 2
    3        > 1
        > 3
    6        > 1
        > 4
    10      > 1
        > 5
    15
    The solution says something about T(n) being of the second degree.
    Then later it states
    1=a+b+c
    3=4a+2b+c
    6=9a+3b+c

    The 1, 3 and 6 part is related to the number of points in the triangle but what does the other part means?
    This seems like a complex sequence at first.
    Can someone help?

    Thanks
    We shall try to find the general expression for the nth term of this sequence.

    1,3,6,10,15,........

    As you may see,

    a_2-a_1=3-1=2

    a_3-a_2=6-3=3

    a_4-a_3=10-6=4

    a_5-a_4=15-10=5

    Generally, a_i-a_{i-1}=i

    Adding all these together you will get, a_i-a_1=2+3+.....+i

    a_i=a_1+2+3+....+i=1+2+.....+i~;~\mbox{ since }a_1=1

    a_i=\frac{i(i+1)}{2}\mbox{ where }i=1,2,3,.......

    Therefore you can find the sum of the first n terms of the sequence by,

    S_n=\sum_{i=1}^{n}a_i=\sum_{i=1}^{n}\frac{i(i+1)}{  2}= \frac{1}{2} \left(\sum_{i=1}^{n}i^{2}+\sum_{i=1}^{n}i\right)= \frac{1}{2} \left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \right)

    S_n=\frac{n(n+1)(n+2)}{6}
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  4. #4
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    Thumbs up Re: Order of Differences, Sequences, Can't find information about it

    Quote Originally Posted by earboth View Post
    2. Therefore the general form of T(n) is
    T(n)=an^2+bn+c

    3. If you plug in n and T(n) into this genral equation you'll get the system of simultaneous equations in (a, b, c):
    \left|\begin{array}{rcl}a+b+c&=&1 \\4a+2b+c&=&3 \\9a+3b+c &=&6 \end{array}\right.
    Solve this system and you should come out with (a, b, c)=\left(\frac12, \frac12, 0\right)

    Hi earboth!
    I understand the most part of your explanation, can you go deeper in the meanings of (a,b,c)?
    I think it may be related to the order of the sequences.
    But I don't understand why a can jump from 1a to 4a
    b is the order of the second column, and c is the order of the third, which would then equal 0

    Another question
    If we would have a sequence with the second column of difference with order of say, 1. The third column of difference would have order zero and we would then considerate T(n) as a cubic expression?


    Quote Originally Posted by Sudharaka View Post
    We shall try to find the general expression for the nth term of this sequence.

    a_i=\frac{i(i+1)}{2}\mbox{ where }i=1,2,3,.......
    Hi Sudharaka! Thanks for your reply.
    I am a little perplexed over how you got a_i=\frac{i(i+1)}{2}

    Is this thought over trial and error?
    I am amazed how this worked, calculating over n=7 we get the right solution.
    Maybe this is a recurrent problem that is easy to calculate?


    Thanks again, it's great to see the way that people that at are good with it, think.
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  5. #5
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    Re: Order of Differences, Sequences, Can't find information about it

    Quote Originally Posted by Zellator View Post
    Hi earboth!
    I understand the most part of your explanation, can you go deeper in the meanings of (a,b,c)?
    I think it may be related to the order of the sequences.
    But I don't understand why a can jump from 1a to 4a
    b is the order of the second column, and c is the order of the third, which would then equal 0

    Another question
    If we would have a sequence with the second column of difference with order of say, 1. The third column of difference would have order zero and we would then considerate T(n) as a cubic expression?




    Hi Sudharaka! Thanks for your reply.
    I am a little perplexed over how you got a_i=\frac{i(i+1)}{2}

    Is this thought over trial and error?
    I am amazed how this worked, calculating over n=7 we get the right solution.
    Maybe this is a recurrent problem that is easy to calculate?


    Thanks again, it's great to see the way that people that at are good with it, think.
    It could be proved that the sum of first i numbers is, \frac{i(i+1)}{2} Please refer, Example 2.3.12: Sum of Squares and Cubes
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  6. #6
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    Re: Order of Differences, Sequences, Can't find information about it

    Quote Originally Posted by Zellator View Post
    Hi earboth!
    I understand the most part of your explanation, can you go deeper in the meanings of (a,b,c)?
    I think it may be related to the order of the sequences.
    But I don't understand why a can jump from 1a to 4a
    b is the order of the second column, and c is the order of the third, which would then equal 0

    Another question
    If we would have a sequence with the second column of difference with order of say, 1. The third column of difference would have order zero and we would then considerate T(n) as a cubic expression? <--- Yes
    .
    1. According to your list you have:
    \left|\begin{array}{rcl}n~&|&~T(n) \\ \hline 1~&|&~T(1)=1 \\ 2~&|& ~ T(2)=3 \\ 3~&|& ~ T(3)=6 \end{array}\right.

    2. a, b and c are the co-efficients of the general term of 2nd degree in n:

    T(n) = a \cdot n^2 + b \cdot n + d

    Now plug in the values of n and T(n) into this general equation and you'll get the system of equations: For each n (and the corresponding T(n)) you'll get one specific equation.

    3. If n = 2 the a \cdot n^2 becomes 4 a ...
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  7. #7
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    Thumbs up Re: Order of Differences, Sequences, Can't find information about it

    Yes! Of course!
    I will not pass things overlooked from now on.

    You said it
    T(n)=an^2+bn+c

    I wasn't seeing we were using n here.

    And thanks Sudharaka, I will put the link to good use.

    Thanks again,
    All the best!
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