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Math Help - Sum of a Series

  1. #1
    Member alexgeek's Avatar
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    Sum of a Series

    I've never come across a series like this, not sure how to sum it:
    <br />
1.2.6 + 2.3.7 + ... + n(n+1)(n+5)<br />

    Maybe something to do with factorials? No idea :/
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  2. #2
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    Quote Originally Posted by alexgeek View Post
    I've never come across a series like this, not sure how to sum it:
    1.2.6 + 2.3.7 + ... + n(n+1)(n+5)
    That notation is so not standard as to to make your question meaningless.
    Please correct it.
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  3. #3
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    Quote Originally Posted by alexgeek View Post
    I've never come across a series like this, not sure how to sum it:
    <br />
1.2.6 + 2.3.7 + ... + n(n+1)(n+5)<br />

    Maybe something to do with factorials? No idea :/
    One way is........

    \displaystyle\sum_{k=1}^nk(k+1)(k+5)=\sum_{k=1}^n\  left[k^3+6k^2+5k\right]

    The sum of cubes is

    \displaystyle\ 1^3+2^3+3^3+....+n^3=(1+2+....+n)^2=\left[\frac{n(n+1)}{2}\right]^2

    The formulae for the "sum of n squares" and the "sum of n terms" can be used for the other two parts.
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  4. #4
    Member alexgeek's Avatar
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    Quote Originally Posted by Plato View Post
    That notation is so not standard as to to make your question meaningless.
    Please correct it.
    I took it straight from the exam paper: http://mathsathawthorn.pbworks.com/f/FP1Jun06.pdf
    Quote Originally Posted by Archie Meade View Post
    One way is........

    \displaystyle\sum_{k=1}^nk(k+1)(k+5)=\sum_{k=1}^n\  left[k^3+6k^2+5k\right]

    The sum of cubes is

    \displaystyle\ 1^3+2^3+3^3+....+n^3=(1+2+....+n)^2=\left[\frac{n(n+1)}{2}\right]^2

    The formulae for the "sum of n squares" and the "sum of n terms" can be used for the other two parts.
    Oh I see, that's kinda of easy really. Thought there'd be some special trickery or something.
    Thanks
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    Quote Originally Posted by alexgeek View Post
    I took it straight from the exam paper: http://mathsathawthorn.pbworks.com/f/FP1Jun06.pdf
    Wasn't the "dot" centered: 1\cdot 2\cdot 4+ 2\cdot3\cdot 7+ \cdot\cdot\cdot?


    Oh I see, that's kinda of easy really. Thought there'd be some special trickery or something.
    Thanks
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Wasn't the "dot" centered: 1\cdot 2\cdot 4+ 2\cdot3\cdot 7+ \cdot\cdot\cdot?
    In some cultures and age groups, the multiplicative dot and decimal point are used the other way around.

    E.g. in Australia, most older generations use a centred dot for a decimal point and a full stop for multiplication, while younger (e.g. mine) generations use the centred dot for multiplication and the full stop for decimals. I assume this is from the introduction of computers - decimals would be used much more frequently than multiplicative dots, so it's easier to use the full stop key for a decimal point.
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  7. #7
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    Code:
    12   42   96   180   300   462...
      30   54   84    120   162
        24   30   36     42
           6    6     6
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  8. #8
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    If you have no formulae to work with, you could analyse patterns...

    k=1,\;2,\;3,\;4,\;

    \sum\;k=1,\;3,\;6,\;10,....

    k^3=1,\;8,\;27,\;64,....

    \sum\;k^3=1,\;9,\;36,\;100,....


    From the pattern, it seems

    \sum\;k^3=\left(\sum\;k\right)^2

    To prove that beyond doubt, you can apply "Proof By Induction".

    Similarly, you can arrive at a formula for the sum of squares...


    k=1,\;2,\;3,\;4,.....

    \sum\;k=1,\;3,\;6,\;10,..

    k^2=1,\;4,\;9,\;16,.....

    \sum\;k^2=1,\;5,\;14,\;30,...

    \displaystyle\frac{\sum\;k^2}{\sum\;k}=\frac{1}{1}  ,\;\frac{5}{3},\;\frac{14}{6},\;\frac{30}{10},....  =\frac{3}{3},\;\frac{5}{3},\;\frac{7}{3},\;\frac{9  }{3},....

    which is an arithmetic sequence pattern of

    \displaystyle\ U_n=a+(n-1)d=1+(n-1)\frac{2}{3}=\frac{3+2n-2}{3}=\frac{2n+1}{3}

    Hence, it seems

    \displaystyle\frac{\sum\;k^2}{\sum\;k}=\frac{2n+1}  {3}\Rightarrow\sum\;k^2=\left[\frac{2n+1}{3}\right]\sum\;k

    Proof By Induction can also verify this...

    Hence all 3 sums can be expressed in terms of \displaystyle\left[\frac{n(n+1)}{2}\right]
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  9. #9
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    Quote Originally Posted by Prove It View Post
    In some cultures and age groups, the multiplicative dot and decimal point are used the other way around.

    E.g. in Australia, most older generations use a centred dot for a decimal point and a full stop for multiplication, while younger (e.g. mine) generations use the centred dot for multiplication and the full stop for decimals. I assume this is from the introduction of computers - decimals would be used much more frequently than multiplicative dots, so it's easier to use the full stop key for a decimal point.
    We seem to use both here,
    <br />
5 \cdot 7, \sin\theta.t, 0.5634<br />
    If I'm writing it I tend to make multiplication dots thicker.

    Quote Originally Posted by Archie Meade View Post
    If you have no formulae to work with, you could analyse patterns...

    k=1,\;2,\;3,\;4,\;

    \sum\;k=1,\;3,\;6,\;10,....

    k^3=1,\;8,\;27,\;64,....

    \sum\;k^3=1,\;9,\;36,\;100,....


    From the pattern, it seems

    \sum\;k^3=\left(\sum\;k\right)^2

    To prove that beyond doubt, you can apply "Proof By Induction".

    Similarly, you can arrive at a formula for the sum of squares...


    k=1,\;2,\;3,\;4,.....

    \sum\;k=1,\;3,\;6,\;10,..

    k^2=1,\;4,\;9,\;16,.....

    \sum\;k^2=1,\;5,\;14,\;30,...

    \displaystyle\frac{\sum\;k^2}{\sum\;k}=\frac{1}{1}  ,\;\frac{5}{3},\;\frac{14}{6},\;\frac{30}{10},....  =\frac{3}{3},\;\frac{5}{3},\;\frac{7}{3},\;\frac{9  }{3},....

    which is an arithmetic sequence pattern of

    \displaystyle\ U_n=a+(n-1)d=1+(n-1)\frac{2}{3}=\frac{3+2n-2}{3}=\frac{2n+1}{3}

    Hence, it seems

    \displaystyle\frac{\sum\;k^2}{\sum\;k}=\frac{2n+1}  {3}\Rightarrow\sum\;k^2=\left[\frac{2n+1}{3}\right]\sum\;k

    Proof By Induction can also verify this...

    Hence all 3 sums can be expressed in terms of \displaystyle\left[\frac{n(n+1)}{2}\right]
    I'm still trying to digest this ha.
    It looks interesting but may be straying too far from the default answers, you're not expected to be at all intuitive in your answers for these papers.
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  10. #10
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    For consistency, I'd encourage getting into the habit of always using a centred dot for multiplication and a full stop for a decimal point.
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  11. #11
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    Quote Originally Posted by alexgeek View Post
    I've never come across a series like this, not sure how to sum it:
    <br />
1.2.6 + 2.3.7 + ... + n(n+1)(n+5)<br />

    Maybe something to do with factorials? No idea :/

    As the product of linear factors I get \frac{n}{4}(n+1)(n+2)(n+3) anyone else agree?
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  12. #12
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    I get

    \displaystyle\left[\frac{n(n+1)}{2}\right]^2+\frac{6n(n+1)}{2}\;\frac{(2n+1)}{3}+\frac{5n(n+  1)}{2}

    =\displaystyle\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+2(2n+1)+5\right]=\frac{n(n+1)}{2}\left[\frac{n^2+n+8n+4+10}{2}\right]

    =\displaystyle\frac{n(n+1)}{2}\left[\frac{n^2+9n+14}{2}\right]=\frac{n(n+1)}{2}\frac{(n+2)(n+7)}{2}
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  13. #13
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    We agree!



    alexgeek, did you get the same?
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  14. #14
    Member alexgeek's Avatar
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    I did indeed
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