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Integrating Binomial expansion

  1. Oct 3, 2006 #1
    Here is the question from the book:

    By integrating the binomial expansion, prove that, for a positive integer n,

    [tex]\frac{2^{n+1} - 1}{n+1} = 1 + \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + ... + \frac{1}{n+1}\binom{n}{n} [/tex]

    So I integrated both sides of the following:

    [tex] (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k [/tex]

    After integrating both sides we get:

    [tex]\frac{(1+x)^{n+1}}{n+1} + C = \sum_{k=0}^n \binom{n}{k} \frac{x^{k+1}}{k+1} [/tex]

    Now my goal makes me feel like plugging in x = 1, which will get us very close to what we want to prove, but that stupid constant is making me slightly off (I think the constant C should be = -1/(n+1) ) But I have no clue how to go about figuring out the value of the constant. Any ideas? Thanks.

    edit... if you do plug in x = 1, you get the following:

    [tex]\frac{2^{n+1}}{n+1} + C = 1 + \frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + ... + \frac{1}{n+1}\binom{n}{n}[/tex]

    Which as I said is very close to what we want to prove, I just can't figure out what we are supposed to do with the constant.
    Last edited: Oct 3, 2006
  2. jcsd
  3. Oct 3, 2006 #2


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    Homework Helper

    Plug in x=0 to solve for the constant.
  4. Oct 3, 2006 #3
    Perfect, thanks!
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