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Math Help - Series Concept

  1. #1
    Senior Member
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    Exclamation Series Concept

    Hey guys

    I'm trying to teach myself about the topic series but can't seem to understand it...

    Esp. these questions:


    1. (x+1)^3 – x^3 = 3x^2 + 3x^ + 1
    ∑n^2 =

    2. (x+1)^4 – x^4 = 4x^3 + 6x^2 + 4x + 1
    ∑n^3=


    Please, culd u please explain how to do these q.s to me??

    Thanx a lot!
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  2. #2
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    Exclamation

    ummm

    For question 1, some of my working out is:

    (x+1)^3 - x^3 = 3x^2 + 3x + 1

    x = 1 2^3 - 1^3 = 3 * 1^2 + 3 * 1 + 1
    x = 2 3^3 - 2^3 = 3* 2^2 + 3 * 2 + 1


    x = n (n+1)^3 - n^3 = 3 * n^3 - n^3 = 3 * n^2 + 3 * n + 1

    (n + 1 - 1)((n+1)^2 + (n+1) + 1^2) = 3En^3 + (3*n(n+1))/2 + n


    -------------------------

    Idno ...somehow we're meant to get En^2 = n/6(n+1)(2n+1) but I don't really noe the rite method to get to it
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  3. #3
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    Hi

    (k+1)^3 - k^3 = 3 k^2 + 3 k + 1

    Sum for k from 0 to n
    (n+1)^3 - 0^3 = 3 \sum_{k=0}^n k^2 + 3 \sum_{k=0}^n k + n+1

    You know that \sum_{k=0}^n k = \frac{n(n+1)}{2}

    Therefore 3 \sum_{k=0}^n k^2 = (n+1)^3 - 3 \frac{n(n+1)}{2} - (n+1)

    Then \sum_{k=0}^n k^2 = \frac{n(n+1)(2n+1)}{6}
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by xwrathbringerx View Post
    ummm

    For question 1, some of my working out is:

    (x+1)^3 - x^3 = 3x^2 + 3x + 1

    x = 1 2^3 - 1^3 = 3 * 1^2 + 3 * 1 + 1
    x = 2 3^3 - 2^3 = 3* 2^2 + 3 * 2 + 1


    x = n (n+1)^3 - n^3 = 3 * n^3 - n^3 = 3 * n^2 + 3 * n + 1

    (n + 1 - 1)((n+1)^2 + (n+1) + 1^2) = 3En^3 + (3*n(n+1))/2 + n


    -------------------------

    Idno ...somehow we're meant to get En^2 = n/6(n+1)(2n+1) but I don't really noe the rite method to get to it
    You have

    {\left( {n + 1} \right)^3} = {n^3} + 3{n^2} + 3n + 1 \Leftrightarrow 3{n^2} = {\left( {n + 1} \right)^3} - {n^3} - 3n - 1

    Now consider the value n = 1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }} \ldots ,{\text{ }}k, i.e.

    3 \cdot {1^2} = {\left( {1 + 1} \right)^3} - {1^3} - 3 \cdot 1 - 1

    3 \cdot {2^2} = {\left( {2 + 1} \right)^3} - {2^3} - 3 \cdot 2 - 1

    3 \cdot {3^2} = {\left( {3 + 1} \right)^3} - {3^3} - 3 \cdot 3 - 1

    \begin{gathered}3 \cdot {4^2} = {\left( {4 + 1} \right)^3} - {4^3} - 3 \cdot 4 - 1 \hfill \\\ldots  \ldots  \ldots  \ldots  \ldots  \ldots  \ldots  \ldots  \ldots  \ldots  \ldots  \hfill \\3 \cdot {k^2} = {\left( {k + 1}\right)^3} - {k^3} - 3 \cdot k - 1 \hfill \\ \end{gathered}

    Now summarizes these values and you will get it

    3 \cdot \sum\limits_{n = 1}^k {{n^2}}  = {\left( {k + 1} \right)^3} - {1^3} - 3 \cdot \sum\limits_{n = 1}^k n  - k = {\left( {k + 1} \right)^3} - \frac{3}{2}k\left( {k + 1} \right) - \left( {k + 1} \right) =

    = \left( {k + 1} \right)\left( {{{\left( {k + 1} \right)}^2} - \frac{3}{2}k - 1} \right) = \left( {k + 1} \right)\left( {{k^2} + \frac{k}{2}} \right) = \frac{{k\left( {k + 1} \right)\left( {2k + 1} \right)}}{2} \Leftrightarrow

    \Leftrightarrow \sum\limits_{n = 1}^k {{n^2}}  = \frac{{k\left( {k + 1} \right)\left( {2k + 1} \right)}}{6}.
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