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Finding the value of combination using integration

  1. Jul 29, 2013 #1
    1. The problem statement, all variables and given/known data
    By using integration, find the value of
    [tex]100C0+\frac{1}{2}100C1+\frac{1}{3}100C2+\frac{1}{4}100C3+...+\frac{1}{101}100C100[/tex]


    2. Relevant equations
    Integration
    Combination


    3. The attempt at a solution
    I don't even have any ideas to start. Can this really be solved by using integration?

    Thanks
     
  2. jcsd
  3. Jul 29, 2013 #2

    haruspex

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    The trick is to turn the sum into a power series, then perform a calculus operation on it which gets rid of those nasty 1/r factors.
     
  4. Jul 29, 2013 #3

    lurflurf

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    recall that for k=1,2,...

    $$\frac{1}{k}=\int_0^1 \! x^{k-1} \, \mathrm{dx} \\ \text{or} \\
    \frac{1}{k}=\int_0^\infty \! e^{-k \, x} \, \mathrm{dx} $$

    use binomial theorem
     
  5. Jul 29, 2013 #4
    I don't think I get both of your hints :redface:

    The question can be written as:

    [tex]\Sigma_{r=0}^{100}~100Cr . \frac{1}{r+1}[/tex]

    Then, by using Maclaurin series:
    [tex]\frac{1}{r+1}=1-r+r^2-r^3+...=\Sigma_{i=0}^{∞}(-r)^i[/tex]

    I am not sure what I am doing and the direction I am heading to....

    Thanks
     
  6. Jul 29, 2013 #5

    haruspex

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    My hint was to introduce a variable, something like (but not exactly):
    [tex]\Sigma_{r=0}^{100}~100Cr . \frac{1}{r+1} s^r[/tex]
     
  7. Jul 30, 2013 #6

    ehild

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    Use binomial theorem to write up 100 (1+x)100. Then integrate it, from 0 to 1. What do you get?


    ehild
     
  8. Jul 30, 2013 #7

    lurflurf

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    my hint was

    $$\sum_{k=0}^{100} {100 \choose k} \, \frac{1}{k+1}=\sum_{k=0}^{100} {100 \choose k} \, \int_0^1 \! x^k \, \mathrm{dx}=\int_0^1\sum_{k=0}^{100} {100 \choose k} x^k \, 1^{100-k} \, \mathrm{dx}$$
    use the binomial theorem
     
  9. Aug 14, 2013 #8
    I am really sorry for taking a long time to reply

    Sorry I still don't get the hint :redface:

    I don't get the last part. Where 1100-k comes from? And also how to use the binomial theorem for your equation? I only know that binomial theorem is used for expanding. In your equation, what is the term that can be expanded using binomial theorem?

    Wow, expanding (1 + x)100 then integrating it from 0 to 1 gives me the same result as calculating

    $$\sum_{k=0}^{100} {100 \choose k} \, \frac{1}{k+1}$$

    So, I can write

    $$\sum_{k=0}^{100} {100 \choose k} \, \frac{1}{k+1}=\int_{0}^{1}{(1+x)^{100}}dx$$

    How can we know such sigma form can be written in simple integral form? Do I have to memorize certain form or expression?

    Thanks
     
  10. Aug 15, 2013 #9

    Dick

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    Just use the binomial theorem on (1+x)^100 and integrate terms. Knowing the binomial theorem is all you have to memorize. Knowing how it relates to to your problem is more of a hunch thing.
     
  11. Aug 15, 2013 #10

    lurflurf

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    I am using the binomial theorem backwards to combine the terms. I insert 1^(100-k) to make the equation exactly fit the binomial theorem. Of course 1^(100-k)=1 so it does not change anything.
    $$\sum_{k=0}^{100} {100 \choose k} x^k \, 1^{100-k}=(1+x)^{100}$$
     
  12. Aug 16, 2013 #11
    Oh ok. Now I can see clearer the connection between stating the question in sigma form and then introducing integration to solve the problem.

    Thanks a lot for all the help I got in this thread :smile:
     
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