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Math Help - formula of a sequence ...

  1. #1
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    formula of a sequence ...

    What is the formula for this sequence:

    1, 4, 8, 13, 19, ...

    I know that the sequence is:

    a_n=\frac {1}{2}(n^2+3n-2)

    The question is (what is the method of finding this answer ?).
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    Re: formula of a sequence ...

    Quote Originally Posted by razemsoft21 View Post
    What is the formula for this sequence:

    1, 4, 8, 13, 19, ...

    I know that the sequence is:

    a_n=\frac {1}{2}(n^2+3n-2)

    The question is (what is the method of finding this answer ?).
    Have a look at this page.
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  3. #3
    Grand Panjandrum
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    Re: formula of a sequence ...

    Quote Originally Posted by razemsoft21 View Post
    What is the formula for this sequence:

    1, 4, 8, 13, 19, ...

    I know that the sequence is:

    a_n=\frac {1}{2}(n^2+3n-2)

    The question is (what is the method of finding this answer ?).
    Construct a difference table, you will find the second differences are constant, so the sequence is a quadratic in the index.

    Then since a quadratic is of the form an^2+bn+c you have:

    \phantom{4}a\phantom{4}+\phantom{2}b+c=1

    4a+2b+c=4

    9a+3b+c=8

    which can be solved for a, b, and c

    CB
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    MHF Contributor chisigma's Avatar
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    Re: formula of a sequence ...

    Quote Originally Posted by razemsoft21 View Post
    What is the formula for this sequence:





    1, 4, 8, 13, 19, ...

    I know that the sequence is:

    a_n=\frac {1}{2}(n^2+3n-2)

    The question is (what is the method of finding this answer ?).
    The sequence is solution of the 'difference equation'...

    a_{n+1}=a_{n}+3+n\ ,\ a_{0}=1 (1)

    ... which can be written as...

    \Delta_{n}=a_{n+1}-a_{n}= 3+n\ ,\ a_{0}=1 (2)

    The 'standard solution' of (1) is...

    a_{n}= a_{0}+\sum_{k=0}^{n-1} \Delta_{k}= a_{0}+ 3 n + \sum_{k=0}^{n-1} k =

    = 1 + 3n + \frac{n\ (n-1)}{2}= \frac{2+5 n +n^{2}}{2} (3)

    Regarding the solving procedure of (1) see the tutorial post on the difference equation in the section 'Discrete Mathematics'...

    Kind regards

    \chi \sigma
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    Re: formula of a sequence ...

    Quote Originally Posted by CaptainBlack View Post
    Construct a difference table, you will find the second differences are constant, so the sequence is a quadratic in the index.

    Then since a quadratic is of the form an^2+bn+c you have:

    \phantom{4}a\phantom{4}+\phantom{2}b+c=1

    4a+2b+c=4

    9a+3b+c=8

    which can be solved for a, b, and c

    CB
    The 3 equations works, but

    how can we Construct a difference table

    the defference is 3,4,5,6, ...

    how do we know that the sequence is a quadratic ?
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    Re: formula of a sequence ...

    Quote Originally Posted by razemsoft21 View Post
    The 3 equations works, but

    how can we Construct a difference table

    the defference is 3,4,5,6, ...

    how do we know that the sequence is a quadratic ?
    What you have there are the first differences, when you take the differences again you get a constant, and you should know that the constant n-th differences mean the original sequence was a polynomial of degree n in the index. Hence as the second differences are constant the original sequence is a quadratic in the index.

    CB
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    Re: formula of a sequence ...

    Quote Originally Posted by chisigma View Post
    The sequence is solution of the 'difference equation'...

    a_{n+1}=a_{n}+3+n\ ,\ a_{0}=1 (1)

    ... which can be written as...

    \Delta_{n}=a_{n+1}-a_{n}= 3+n\ ,\ a_{0}=1 (2)

    The 'standard solution' of (1) is...

    a_{n}= a_{0}+\sum_{k=0}^{n-1} \Delta_{k}= a_{0}+ 3 n + \sum_{k=0}^{n-1} k =

    = 1 + 3n + \frac{n\ (n-1)}{2}= \frac{2+5 n +n^{2}}{2} (3)

    Regarding the solving procedure of (1) see the tutorial post on the difference equation in the section 'Discrete Mathematics'...

    Kind regards

    \chi \sigma
    Equation (3) does not match mine
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    Re: formula of a sequence ...

    Quote Originally Posted by CaptainBlack View Post
    What you have there are the first differences, when you take the differences again you get a constant, and you should know that the constant n-th differences mean the original sequence was a polynomial of degree n in the index. Hence as the second differences are constant the original sequence is a quadratic in the index.

    CB
    uderstood, thanks

    finaly how can we form the 3 equations?
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    Re: formula of a sequence ...

    Quote Originally Posted by razemsoft21 View Post
    Equation (3) does not match mine
    I do not follow his workings either.
    Here is my way.
    a_1=1 and if n\ge 2 then a_n=a_{n-1}+(n+1).
    Thus
    \begin{gathered}  a_n  \hfill \\   = a_1  + \sum\limits_{k = 2}^n {\left( {k + 1} \right)}  \hfill \\   = 1 + \frac{{n\left( {n + 1} \right)}}{2} - 1 + \left( {n - 1} \right) \hfill \\   = \frac{1}{2}\left( {n^2  + 3n - 2} \right) \hfill \\ \end{gathered}
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    Re: formula of a sequence ...

    Quote Originally Posted by razemsoft21 View Post
    uderstood, thanks

    finaly how can we form the 3 equations?
    Let P(n)=an^2+bn+c, then:

    P(1)=1

    from the first term in the sequence, and:

    P(2)=4

    from the 2nd term, and:

    P(3)=8

    from the third.

    These are all the terms that need be used since the three simultaneous equations will have a unique solution (and it will be the same for three equations formed from any three distinct terms in the sequence).

    Also Newton has a systematic way of doing this from the difference table that you may have covered in class.

    CB
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  11. #11
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    Re: formula of a sequence ...

    Thanks for all of you

    another thank for Soroban

    the 2 ways are work and understood.
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    MHF Contributor chisigma's Avatar
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    Re: formula of a sequence ...

    Quote Originally Posted by razemsoft21 View Post
    Equation (3) does not match mine
    It is a question of 'initial index'... if the initial index is 0, then the difference equation is...

    a_{n+1}=a_{n}+3+n,\ a_{0}=1 (1)

    ... and its solution is...

    a_{n}= \frac{2+5 n +n^{2}}{2} (2)

    ... if the initial index is 1, then the difference equation is...

    a_{n+1}=a_{n}+2+n,\ a_{1}=1 (3)

    ... and its solution is...

    a_{n}= \frac{-2+3 n +n^{2}}{2} (4)

    In any case You have as solution the sequence 1,4,8,13,19,...

    Kind regards

    \chi \sigma
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  13. #13
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    Re: formula of a sequence ...

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