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Limit of summation notation

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data

    http://desmond.imageshack.us/Himg100/scaled.php?server=100&filename=img20120327195119.jpg&res=medium [Broken]

    2. Relevant equations



    3. The attempt at a solution

    I just plugged in ∞ for n

    [2+[itex]\frac{3}{∞}[/itex]]2 ([itex]\frac{3}{∞}[/itex]) =

    [2+0]2 (0) = 0


    Did I do the problem correctly? I might need a refresher on summation notations.

    Here are the multiple choice answers:

    a. 0
    b. 1
    c. 4
    d. 39
    e. 125
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 27, 2012 #2
    Seems right to me.
     
  4. Mar 27, 2012 #3

    Dick

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    It's not right at all. You can't plug n=infinity into that. It looks like a Riemann sum approximation to an integral to me. None of the multiple choice answers that you've shown are correct either.
     
  5. Mar 27, 2012 #4

    I couldn't fit all of the choices into the frame.

    The choices:

    a. 0
    b. 1
    c. 4
    d. 39
    e. 125
     
  6. Mar 27, 2012 #5

    Dick

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    I'm not going to pick an answer for you. Show me how to get it. Put x=k/n and express that as a limiting sum for a Riemann integral over x.
     
  7. Mar 27, 2012 #6
    Okay , I think I did it.


    (3/n) = ΔX
    ΔX = (b-a)/n

    so, b=3 ; a=0

    xi = a + [i(b-a)]/n

    xi = [0 + (3k)/n + 2]2

    f(xi) = (x+2)2


    The integral would be [itex]^{3}_{0}[/itex]∫(x+2)2
     
  8. Mar 28, 2012 #7

    Ray Vickson

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    One of the listed answers is correct.

    RGV
     
  9. Mar 28, 2012 #8

    Dick

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    Right.
     
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