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Finding the sum of

  1. Jan 15, 2006 #1
    finding the sum of....

    1*2+2*3+3*4+...+n(n+1)

    i need to find to what it's equal?

    i tried doing various ways:

    2*(1+3)+4*(3+5)+6(5+7)+...

    which means an even number times the sum of two odd numbers
    so let's say:
    a=2m
    b=2m-1

    a(b+3)+a(b+5)+a(b+7)...+n(n+1)
    ab*n+a(3+5+7...+2n-1)*(n-1)=
    a{[bn+(n-1)(3+5+7+...2n-1)]}

    when 3+5+7+...(2n-1)=(3+s_n)n/2
    when s_n=3+n(2n-1-3)=2n^2-4n+3
    so (3+s_n)n/2=(2n^2-4n+6)n/2=n^3-2n^2+3n

    here i pretty much given up, can someone help me.
     
  2. jcsd
  3. Jan 15, 2006 #2

    benorin

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  4. Jan 15, 2006 #3

    benorin

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    In particular,

    [tex]\sum_{k=1}^{n} k(k+1) = \frac{n(n+1+1)!}{(1+2)n!} = \frac{n(n+1)(n+2)}{3}[/tex]
     
  5. Jan 15, 2006 #4
    i think you jumped there too quickly, here:
    sigma[k=1 to n](k+j)!/(k-1)!=n(n+j+1)!/(j+2)n!

    the sigma equals:
    (1+j)!/0!+(2+j)/1!!+...+(n+j)!/(n-1)!

    so where's the thing in between?
     
  6. Jan 15, 2006 #5

    benorin

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    The proof ? Oh, just I used Maple :wink: .
     
    Last edited: Jan 15, 2006
  7. Jan 15, 2006 #6
    so does someone know how to prove it, or should i use my bare hands. (-:
     
  8. Jan 15, 2006 #7

    George Jones

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    Now that you know the final answer, the result should be easy to prove by mathematical induction.

    Also...
    This method works! (For an even number of terms.)

    [tex]
    \begin{equation*}
    \begin{split}
    1*2 + 2*3 + 3*4 + 4*5 + 5*6 + 6*7 + 7*8 + 8*9 + \cdots &= 2 \left( 1 + 3 \right) + 4 \left( 3 + 5 \right) + 6 \left( 5 + 7 \right) + 8 \left( 7 + 9 \right) + \cdots\\
    &= 8 + 32 + 72 + 128 + \cdots\\
    &= 8 \left( 1 + 4 + 9 + 16 + \cdots \right)\\
    &= 8 \left( 1 + 2^{2} + 3^{2} + 4^{2} + \cdots \right)\\
    \end{split}
    \end{equation*}
    [/tex]

    The stuff in the brackets is a known series, but this is not a proof. It does, however, suggest the following more formal method. Let [itex]n = 2m[/itex], so that

    [tex]\sum_{k=1}^{2m} k \left( k + 1 \right) = \sum_{k=1}^{2m} a_{k},[/tex]

    where [itex]a_{k} = k \left( k + 1 \right)[/itex]. Define a series by [itex]b_{1} = a_{1} + a_{2}[/itex], [itex]b_{2} = a_{3} + a_{4}[/itex], etc. In other words, [itex]b_{j} = a_{2j - 1} + a_{2j}[/itex].

    Now,

    [tex]
    \begin{equation*}
    \begin{split}
    \sum_{k=1}^{2m} k \left( k + 1 \right) &= \sum_{k=1}^{2m} a_{k}\\
    &= \sum_{j=1}^{m} b_{j}\\
    &= \sum_{j=1}^{m} \left[ a_{2j-1} + a_{2j} \right] .\\
    \end{split}
    \end{equation*}
    [/tex]

    What do you get when you sub in for the a's?

    Regards,
    George
     
    Last edited: Jan 15, 2006
  9. Jan 15, 2006 #8

    HallsofIvy

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    I would think that it would be simpler to write
    [tex]\begin{equation*}\begin{split}\sum_{k=1}^{n} k \left( k + 1 \right) \\&= \sum_{k=1}^{n}(k^2+ k)\\&= \sum_{k=1}^{n} k^2+ \sum_{k=1}^{n} k\end{split}\end{equation*}[/tex]
    And those two sums are well known.
     
  10. Jan 15, 2006 #9

    George Jones

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    Yikes! :blushing:

    This exemplifies something I often tell people: take care when replacing symbols by numbers, because this sometimes complicates the issue at hand.

    Mea culpa.

    Regards,
    George
     
  11. Jan 16, 2006 #10
    thank, i believe this is much easier from what i had thought before.

    the first k^2 sum is of what galileo find, and the second one is just plane arithmatic progression.
     
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