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Binomial series with coeficients in arithmetic progression

  1. Sep 25, 2014 #1
    1. The problem statement, all variables and given/known data
    The binomial expansion of [itex](1+x)^n[/itex], n is a positive integer, may be written in the form

    [itex](1+x)^{n} = 1+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+...c_{r}x^{r}+...
    [/itex]

    Show that , if [itex]c_{s-1}[/itex], [itex]c_{s}[/itex] and [itex]c_{s+1} [/itex] are in arithmetic progression then [itex](n-2s)^{2} =n+2[/itex]

    2. Relevant equations



    3. The attempt at a solution
    "[itex]c_{s-1}[/itex], [itex]c_{s}[/itex] and [itex]c_{s+1} [/itex] are in arithmetic progression" infers that
    [itex]\frac{c_{s-1} +c_{s+1}}{2}=c_{s}\\
    \frac{\binom{n}{s-1} + \binom{n}{s+1}}{2}=\binom{n}{s} \\
    \frac{n!}{2(n-s+1)!(s-1)!} + \frac{n!}{2(n-s-1)!(s+1)!} = \frac{n!}{(n-s)!s!}\\
    \frac{(n-s)!s!}{(n-s+1)!(s-1)!} + \frac{(n-s)!s!}{(n-s-1)!(s+1)!} = 2\\
    \frac{s}{n-s+1} + \frac{n-s}{s+1}= 2\\
    \frac{n-s}{s+1} = 2- \frac{s}{n-s+1}\\
    n+2= (s+1)(2- \frac{s}{n-s+1})+s+2\\
    n+2= 2(s+1)- \frac{s(s+1)}{n-s+1}+s+2
    [/itex]
    At this point it's getting messy, so I set s to 1 and n to 2 and get n+2 = 6 as opposed to the 0 I get when I do the same susubstitution for [itex](n-2s)^{2}[/itex]
    Can someone please point out what I have done wrong.
     
  2. jcsd
  3. Sep 25, 2014 #2

    haruspex

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    It's no use checking the equation by using a combination of values that is not actually a solution. n=7, s=2 is a solution, and your last equation survives that test.
    Go back to your 5th equation (the one with just =2 on the right hand side) and just multiply out the denominators.
     
  4. Sep 25, 2014 #3
    OK, yes, pretty fundamental misunderstanding then. Thanks for pointing that out.
     
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