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Math Help - Partial Sums of a Series

  1. #1
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    Partial Sums of a Series

    Hey folks, been an awful long time since I've been working with series, could definitely use a push with this problem.

    The series obtained from the sequence a_{n}=3n+2 , is the restriction to the integers of a polynomial function y=f(x). Construct the partial sums of such series and use them to identify the function f(x).

    I started by writing the series in sigma notation,

    \sum_{i=1}^n (3i+2)

    And to construct the partial sums, we can split up the total sum into:

    \sum_{i=1}^n (3i+2)

    <br />
=\sum_{i=1}^n (3i) + \sum_{i=1}^n (2)

    =3 \sum_{i=1}^n (i) + 2

    Right?

    I want to say that f(x)=3x+2 which makes sense as it would be a_{n}=3n+2 if restricted to the integers, but I'm not sure if I've found my correct partial sum?

    Thanks for your time!
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  2. #2
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    Quote Originally Posted by Kasper View Post
    Hey folks, been an awful long time since I've been working with series, could definitely use a push with this problem.

    The series obtained from the sequence a_{n}=3n+2 , is the restriction to the integers of a polynomial function y=f(x). Construct the partial sums of such series and use them to identify the function f(x).

    I started by writing the series in sigma notation,

    \sum_{i=1}^n (3i+2)

    And to construct the partial sums, we can split up the total sum into:

    \sum_{i=1}^n (3i+2)

    <br />
=\sum_{i=1}^n (3i) + \sum_{i=1}^n (2)

    =3 \sum_{i=1}^n (i) + 2

    Right?

    I want to say that f(x)=3x+2 which makes sense as it would be a_{n}=3n+2 if restricted to the integers, but I'm not sure if I've found my correct partial sum?

    Thanks for your time!
    It would actually be

    =\sum_{i=1}^n (3i) + \sum_{i=1}^n (2)

     = 3\left(\sum_{i = 1}^n {i}\right) + 2n

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

     = \frac{3n}{2} + \frac{3n^2}{2} + 2n

     = \frac{3n^2}{2} + \frac{7n}{2}.
    Last edited by Prove It; September 27th 2009 at 09:08 AM.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    It would actually be

    =\sum_{i=1}^n (3i) + \sum_{i=1}^n (2)

     = 3\sum_{i = 1}^n (i) + 2n

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

     = \frac{3}{2} + \frac{3}{2}n + 2n

     = \frac{3}{2} + \frac{7}{2}n.
    Ah, I missed how the sum to n terms of 2 would be 2n, of course. :/

    I am however still confused on how \sum_{i=1}^n (3i) = \frac{1}{2}(1+n). Series have definitely never been a strong point for me.
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  4. #4
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    Quote Originally Posted by Kasper View Post
    Ah, I missed how the sum to n terms of 2 would be 2n, of course. :/

    I am however still confused on how \sum_{i=1}^n (3i) = \frac{1}{2}(1+n). Series have definitely never been a strong point for me.
    Note, I edited the above post.

    We have

    \sum_{i = 1}^n {(3i)} = 3\left(\sum_{i = 1}^n {i}\right)


    Notice that \sum_{i = 1}^n {i} = 1 + 2 + 3 + \dots + n.


    This is an arithmetic series with t_1 = 1, t_n = n, and common difference d = 1.

    The sum of an arithmetic sequence can be found using the formula

    \sum_{i = 1}^n{i} = \frac{n}{2}(t_1 + t_n)

     = \frac{n}{2}(1 + n).


    Thus 3\left(\sum_{i = 1}^n{i}\right) = \frac{3n}{2}(1 + n).
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Note, I edited the above post.

    We have

    \sum_{i = 1}^n {(3i)} = 3\left(\sum_{i = 1}^n {i}\right)


    Notice that \sum_{i = 1}^n {i} = 1 + 2 + 3 + \dots + n.


    This is an arithmetic series with t_1 = 1, t_n = n, and common difference d = 1.

    The sum of an arithmetic sequence can be found using the formula

    \sum_{i = 1}^n{i} = \frac{n}{2}(t_1 + t_n)

     = \frac{n}{2}(1 + n).


    Thus 3\left(\sum_{i = 1}^n{i}\right) = \frac{3n}{2}(1 + n).
    Oh! That makes perfect sense, I'll have to keep that formula handy, thanks so much!
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