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Homework Help: Simplifying Sigma Notation

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm actually asked to calculate the area under the curve 5x + x2 over the interval [0,1] using Riemann Sums. I found the formula for the Riemann sum over the interval, it being the following:

    [itex]\sum[/itex][itex]^{n}_{k=1}[/itex] ([itex]\frac{5k}{n}[/itex]+[itex]\frac{k^{2}}{n^{2}}[/itex])([itex]\frac{1}{n}[/itex])

    However, I am asked to simplify the sigma notation to find the sum in terms of n only, which is where I'm currently stuck.

    2. Relevant equations

    None, I believe. Except perhaps the sum formulas for positive integers

    3. The attempt at a solution

    I actually just went ahead and multiplied through and separated each term like so

    [itex]\frac{5}{n^{2}}[/itex][itex]\sum[/itex][itex]^{n}_{k=1}[/itex] k +[itex]\frac{1}{n^{3}}[/itex][itex]\sum[/itex][itex]^{n}_{k=1}[/itex] k[itex]^{2}[/itex]

    I, however, was left with something completely different from the expected answers given. Am I going through this correctly?
  2. jcsd
  3. Apr 12, 2012 #2
    Yes, you are on the right track. Now use a closed-form expression for those two sums to get something only in terms of n; then you can take the limit as n goes to infinity.
  4. Apr 12, 2012 #3
    Okay, I figured out what I was doing wrong after going through my work - I neglected to multiply part of the numerator of the first sum by 3n, thus giving me something different. So after multiplying through, the end result should be:

    [itex]\frac{15n^{3}+15n^{2}}{6n^{3}}[/itex] + [itex]\frac{2n^{3}+3n^{2}+n}{6n^{3}}[/itex]

    This should simplify to:


    This could further be simplified by removing [itex]\frac{17n^{3}}{6n^{3}}[/itex] to get

    [itex]\frac{17}{6}[/itex] + [itex]\frac{18n+1}{6n^{2}}[/itex], where an n is distributed out of the second term.

    Then the limit should be [itex]\frac{17}{6}[/itex] as n approaches infinity since the second term is equal to zero. :D
  5. Apr 12, 2012 #4


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    Sure, that's it.
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