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Math Help - Sums of Series?

  1. #1
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    Question Sums of Series?

    Given that

    S(N)=

    N
    Σ...(2n-1)^3
    n=1

    Show that
    S(N)=N^2(2N^2 - 1)

    I tried expanding it and used the following formulas:
    Σn=n(n+1)/2
    Σn^2=n(n+1)(2n+1)/6
    Σn^3=n^2(n+1)^2/4

    But it doesn't give the answer.

    I also tried substituting (2n-1) directly into the last formula but it also doesn't give the answer.

    Please help,
    Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Sums of Series?

    Quote Originally Posted by Srimadeva1996 View Post
    Given that

    S(N)=

    N
    Σ...(2n-1)^3
    n=1

    Show that
    S(N)=N^2(2N^2 - 1)

    I tried expanding it and used the following formulas:
    Σn=n(n+1)/2
    Σn^2=n(n+1)(2n+1)/6
    Σn^3=n^2(n+1)^2/4

    But it doesn't give the answer.

    I also tried substituting (2n-1) directly into the last formula but it also doesn't give the answer.

    Please help,
    Thanks
    Starting with the general formula...

    S_{n}= \sum_{k=1}^{n} k^{3}= \frac{n^{2}\ (n+1)^{2}}{4} (1)

    ... we derive from (1)...

    S_{2n-1}= \sum_{k=1}^{2n-1} k^{3}= n^{2}\ (2n-1)^{2} (2)

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

    .. so that is...

    \sum_{k=1}^{n} (2k-1)^{3} = S_{2n-1} - S_{2n} (4)

    The last details are left to You...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor
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    Re: Sums of Series?

    Quote Originally Posted by Srimadeva1996 View Post
    Given that

    S(N)=

    N
    Σ...(2n-1)^3
    n=1

    Show that
    S(N)=N^2(2N^2 - 1)

    I tried expanding it and used the following formulas:
    Σn=n(n+1)/2
    Σn^2=n(n+1)(2n+1)/6
    Σn^3=n^2(n+1)^2/4

    But it doesn't give the answer.

    I also tried substituting (2n-1) directly into the last formula but it also doesn't give the answer.

    Please help,
    Thanks
    \sum_{n=1}^ N n^3=\frac{N^2(N+1)^2}{4}


    \sum_{n=1}^N (2n-1)^3=1^3+3^3+5^3+.......+(2N-1)^3

    =1^3+2^3+3^3+....+(2N-2)^3+(2N-1)^3-\left[2^3+4^3+....(2N-2)^3\right]

    =\sum_{n=1}^{2N-1} n^3-\left[2^3+4^3+6^3+....+(2N-2)^3\right]

    =\sum_{n=1}^{2N-1} n^3-2^3\left[1^3+2^3+3^3+....+(N-1)^3\right]

    =\sum_{n=1}^{2N-1} n^3-2^3\sum_{n=1}^{N-1} n^3

    and now you may use your formula for the sum of cubes.
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  4. #4
    Grand Panjandrum
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    someplace
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    Re: Sums of Series?

    Quote Originally Posted by Srimadeva1996 View Post
    Given that

    S(N)=

    N
    Σ...(2n-1)^3
    n=1

    Show that
    S(N)=N^2(2N^2 - 1)

    I tried expanding it and used the following formulas:
    Σn=n(n+1)/2
    Σn^2=n(n+1)(2n+1)/6
    Σn^3=n^2(n+1)^2/4

    But it doesn't give the answer.

    I also tried substituting (2n-1) directly into the last formula but it also doesn't give the answer.

    Please help,
    Thanks
    Have you considered using induction?

    CB
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  5. #5
    Super Member

    Joined
    May 2006
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    Lexington, MA (USA)
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    Re: Sums of Series?

    Hello, Srimadeva1996!

    \text{Given that: }\:S(N) \;=\;\sum^N_{n=1}(2n-1)^3

    \text{Show that: }\;S(N)\:=\:N^2(2N^2 - 1)


    I tried expanding it and used the following formulas:

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

    . . \sum n^2 \:=\:\frac{n(n+1)(2n+1)}{6}

    . . \sum n^3 \:=\:\frac{n^2(n+1)^2}{4}

    But it doesn't give the answer. .
    It should!

    \sum(2n-1)^3 \;=\;\sum(8n^3 - 12n^2 + 6n - 1)

    . . . . . . . . . =\;8\left(\sum n^3\right) - 12\left(\sum n^2\right) + 6\left(\sum n\right) - \left(\sum 1\right)

    . . . . . . . . . =\;8\cdot\frac{N^2(N+1)^2}{4} - 12\cdot\frac{N(N+1)(2N+1)}{6} + 6\cdot\frac{N(N+1)}{2} - N

    . . . . . . . . . =\;2N^2(N+1)^2 - 2N(N+1)(2N+1) + 3N(N+1) - N

    . . . . . . . . . =\;2N^2(N^2+2N+1) - 2N(2N^2 + 3N + 1) + 3N^2 + 3N - N

    . . . . . . . . . =\;2N^4 + 4N^3 + 2N^2 - 4N^3 - 6N^2 - 2N + 3N^2 + 3N - N

    . . . . . . . . . =\;2N^4 - N^2

    . . . . . . . . . =\;N^2(2N^2-1)
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