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How to solve this Binomial Theorem problem ?

  1. Aug 16, 2009 #1
    1. The problem statement, all variables and given/known data

    106g8jd.jpg

    2. Relevant equations

    Formula => C(n,r) or nCr =n!/r!(n-r)! & the basic Binomial Theorem formula.

    *Answer mentioned in book = nx

    3. The attempt at a solution

    The LHS should be (x+y)n & the given question is its expansion only if that 'r' is not multiplied in the question. I also know how to add using the summation feature but I am not able to get x+y in the expansion (so I could substitute the value of 1). If I substitute 1 on the LHS I get => 1n.
     
    Last edited: Aug 16, 2009
  2. jcsd
  3. Aug 16, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi gaganspidey! Welcome to PF! :smile:
    hmm … when you see the trick, you're going to kick yourself about how simple it is. :biggrin:

    Hint: the sum of n elements is x times n … think "average" … :rolleyes:

    (and everyone else, please no more hints until he's had a chance to reply! :wink:)
     
  4. Aug 17, 2009 #3
    It might sound strange but I still don't get it. From your hint what I've been able to understand is that there are n elements in the expansion because r ranges from 0 to n, but what has x got to do with it, I mean isn't it something like this -

    0 + nxyn-1 +n(n-1)x2yn-2 + .... + xn

    So how do we solve it to get the answer.

    So sorry for being silly, I've a feeling its really easy but I am just not able to get it :frown:
     
    Last edited: Aug 17, 2009
  5. Aug 17, 2009 #4

    tiny-tim

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    Sorry … perhaps that was a bit too cryptic :redface:

    Try this: imagine you do the same thing n times (eg, try to hit a target), with independent results, and the probability of success each time is x:

    what does ∑ rnCrxryn-r represent? :smile:
     
  6. Aug 18, 2009 #5
    So its means we don't actually have to calculate anything here, we just have to assume its average, like you said earlier,i.e, :-

    = 0 + nxyn-1 +n(n-1)x2yn-2 + .... + xn <-----(the first actual step)
    = x + x + x + x + .... + x upto n terms <-----(assuming we get x by solving each term on average)
    = xn

    But then why have the given value 'x+y=1' not been used, I think it tells us that the values of 'x' & 'y' are less than 1. And why do we take the average of each term as 'x' and not as any other term such as 'y' or 'r'
     
  7. Aug 19, 2009 #6

    tiny-tim

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    ∑ rnCrxryn-r represents the expected value of the number of successes (after n independent tries).

    But if you haven't done "expected values", try this:

    What is ∑ (r/nx) nCrxryn-r ? :smile:

    (expand the nCr)
     
  8. Aug 19, 2009 #7
    Unfortunately I'm not aware of "expected value" concept.
    Here's the expansion that I've done :
    = 0 + (1/nx) n(n-1)!/(n-1)! xyn-1 + (2/nx) n(n-1)(n-2)!/2!(n-2)! x2yn-2 + ... + (n/nx) nCnxnyn-n <------(substituting r=0,1,2...n)
    = yn-1 + (n-1)xyn-2 + ... + xn-1
     
  9. Aug 19, 2009 #8

    jambaugh

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    I find that for general problems like this you need to manipulate the binomial coefficient expressed as a ratio of factorials(absorb the r term into it).

    Note:
    [tex]r! = r\cdot (r-1)! = r\cdot (r-1)\cdot (r-2)! ...[/tex]

    Thus:

    [tex]\sum_{r=0}^{n} r \mathbf{C}^n_r x^r y^{n-r}=\sum_{0}^{n} r \cdot\frac{n!}{r!(n-r)!} x^r y^{n-r}[/tex]

    [tex]=\sum_{r=0}^{n} r \cdot\frac{n!}{r\cdot(r-1)!(n-r)!} x^r y^{n-r}[/tex]

    Cancel the r's and then manipulate until you get a different binomial coefficient. Then see if you can factor out x's or y's from the sum to get another different binomial expansion. Then pay close attention to the end terms on the summation. See if you can manipulate it into something plus something times another binomial expansion sum.

    Note that you can recursively solve this with higher powers of r since:

    [tex]\sum_{r=0}^{n} r^2 \frac{n!}{r!(n-r)!} x^r y^{n-r}[/tex]

    [tex]\sum_{r=0}^{n} r(r-1) \frac{n!}{r!(n-r)!} x^r y^{n-r}-\sum_{0}^{n} r \frac{n!}{r!(n-r)!} x^r y^{n-r}[/tex]

    [EDIT: I forgot to write r=0 instead of 0 in the summation notation. Fixed now.]
     
    Last edited: Aug 19, 2009
  10. Aug 19, 2009 #9

    tiny-tim

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    hold it! what's going on here? :confused:

    you're getting really confused …

    n(n-1)(n-2)/2!(n-2)! isn't nC2

    (I assume you meant that, without the extra "!")

    you either mean n(n-1)/2!, or you mean n!/2!(n-2)!

    Try again, multiplying by r/nx, and just using n!/r!(n-r)! (and without breaking it down any further) :smile:
     
  11. Aug 19, 2009 #10
    Ok, you're on the right track.

    What don't you use the sum of geometric series?

    [tex]a + ar + a r^2 + ar^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^n}{1-r},[/tex]

    Regards.
     
  12. Aug 19, 2009 #11
    I think this is an excellent way of doing this! Since neither n nor x are involved in the summation index, you can factor both of them out. And what's left can be manipulated into something much nicer.

    What do you mean by this? Apparently I'm a bit out of it and I'm not sure what you're suggesting. Thanks!
     
  13. Aug 19, 2009 #12

    Gib Z

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    Alternatively, we could notice that the sum is the derivative of the usual binomial expansion multiplied by x.
     
  14. Aug 20, 2009 #13
    Yes actually I expanded it a little bit further, but when applying the formula only, I get :

    = 0 + (1/nx) n!/(n-1)! xyn-1 + (2/nx) n!/2!(n-2)! x2yn-2 + ... + (n/nx) nCnxnyn-n <------(substituting r=0,1,2...n)
    = yn-1 + (n-1)xyn-2 + ... + xn-1

    which is the same result.
     
  15. Aug 20, 2009 #14
    I wonder how I could apply this formula because we need a geometric progression for it,ie, a2/a1=a3/a2 whereas I am not getting LHS = RHS.
     
  16. Aug 20, 2009 #15
    Hmm...got to give it a try.
     
  17. Aug 20, 2009 #16
    Sorry, I still have to learn derivation/differentiation.
     
  18. Aug 20, 2009 #17

    tiny-tim

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    Nooo …

    you're trying to guess what it is from what the first couple of terms look like …

    that's a recipe for disaster …

    just deal with one (general) term: what is (r/nx)nCr ?

    (this is the same as jambaugh's :smile: suggestion …)
    (and ignore Дьявол's suggestion)
     
  19. Aug 20, 2009 #18

    Gib Z

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    Ahh sorry, I forgot it was in the pre calc forums. Pity, I thought it was a pretty short and simple solution =[

    EDIT: I guess it depends on what you can see from inspection. I mean, if you can see that your series is the expansion of [itex]nx (x+y)^{n-1}[/itex] then that would be great lol! But obviously it needs practice for these kind of things.
     
  20. Aug 20, 2009 #19

    jambaugh

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    I worked it out and there are a couple of "tricks" trying from the start (such as adding and subtracting 1 and grouping to get a particular form.)

    I suggest you also give the binomial expansion of:
    [tex](x+y)^{n-1}[/tex]
    so you can work from both ends.

    Also it is quite helpful to pick some small values of n say 3, 4 and 5 and working through theses examples explicitly to see what's going on in general. (Good technique in any such problem!)
     
  21. Aug 21, 2009 #20
    This went over my head :redface:
     
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