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Question regarding summation of series

  1. Apr 5, 2005 #1
    If [tex]\sum^{n}_{r=1} u_r =3n^2 +4n [/tex], what is [tex] \sum^{n-1}_{r=1}u_r [/tex] ?

    I know that [tex] \sum^{n-1}_{r=1}u_r [/tex] is equals to [tex]\sum^{n}_{r=1} u_r =3n^2 +4n - u_n [/tex] but the answer given is [tex] 3n^2-2n-1[/tex]. How do i express it in that way?

    thanks alot.
     
  2. jcsd
  3. Apr 5, 2005 #2

    dextercioby

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    Use this

    [tex] \sum_{r=1}^{n} r =\frac{n(n+1)}{2} [/tex]

    Daniel.
     
  4. Apr 5, 2005 #3
    Substitute n with n-1:

    [tex]\sum^{n}_{r=1} u_r =3n^2 +4n [/tex]
    becomes
    [tex]\sum^{n-1}_{r=1} u_r =3(n-1)^2 +4(n-1)
    =3(n^2 -2n +1) +4(n -1)
    =3n^2 -2n -1[/tex]
     
  5. Apr 5, 2005 #4

    dextercioby

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    Heh,again,thank god someone came with a more intuitive method...:smile:

    I would have found the general term [itex] u_{r}=6r+1 [/itex] ...

    Daniel.
     
  6. Apr 5, 2005 #5
    hey thanks daniel. I've used that to find out that [tex] sum^{n}_{r=1} u_r [/tex] is an arithmetic progression with difference 6 and first term 7. So, i went on from there to find out the answer.

    to whkoh: i don't understand your second step, [tex]\sum^{n-1}_{r=1} u_r =3(n-1)^2 +4(n-1)
    =3(n^2 -2n +1) +4(n -1)
    =3n^2 -2n -1[/tex].

    would you mind explaining it to me? thanks alot.
     
  7. Apr 5, 2005 #6
    The second step is expanding out so that it becomes
    [tex]3(n-1)^2+4(n-1)
    =3(n^2 -2n+1)+4n-4
    =3n^2 -6n+4n+3-4
    =3n^2 -2n-1
    [/tex]
     
  8. Apr 5, 2005 #7
    I don't understand how you find the general term [tex]U_r[/tex]?
     
  9. Apr 5, 2005 #8

    dextercioby

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    [tex] 6 \sum_{r=1}^{n} r =6\frac{n(n+1)}{2}=3n^{2}+3n [/tex] (1)

    U have [tex] 3n^{2}+4n [/tex],so you need another "n"...That can be gotten noticing that

    [tex] \sum_{r=1}^{n} 1 = n [/tex] (2)

    Add (1) & (2) and u'll get

    [tex] \sum_{r=1}^{n} \left(6r+1\right) = 3n^{2}+4n [/tex] (3)

    The conclusion is simple.U found the general term,so for a summation till [itex] n-1 [/itex] simply subtract the general term from the sum till [itex] n [/itex]...

    Daniel.

    P.S.Your method is simpler,mine is due to intuition only.
     
  10. Apr 5, 2005 #9
    ahhh thanks alot. I finally understood..
     
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