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Homework Help: Simplification Difficulties

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data

    I want to simplify the following (these are to do with the ratio test in series):

    a) [tex]\frac{(n+1)!}{(2n+1)}/\frac{n!}{2n-1}[/tex]

    b) [tex]\frac{2^{2n+1}}{(2n+1)!} / \frac{2^{2n-1}}{(2n-1)!}[/tex]

    3. The attempt at a solution

    a) [tex]\frac{(n+1)!}{(2n+1)}/\frac{n!}{2n-1}[/tex]

    [tex]\frac{(n+1)!}{(2n+1)} \times \frac{(2n-1)}{n!}[/tex]

    So, we devide up the factorial terms: [tex]\frac{n \times n+1}{n} = n+1[/tex]

    and deviding the rest:


    Therefore the whole thing should simplify to: -(n+1) But the book says it must be [tex]\frac{n+1}{2n+1}[/tex], what's wrong?

    b) [tex]\frac{2^{2n+1}}{(2n+1)!} / \frac{2^{2n-1}}{(2n-1)!}[/tex]

    [tex]\frac{2^{2n+1}}{(2n+1)!} \times \frac{2^{2n-1}}{(2n-1)!}[/tex]

    [tex]\frac{4n . 2}{4n . 1/2} = 4[/tex]

    (2n-1)!/(2n+1)! = [tex]\frac{(2n)(2n-1)}{(2n)(2n+1)}[/tex]

    So based on what I've done my answer has to be -4 but the real answer has to be:

    Can anyone please help me with this simplifications...
  2. jcsd
  3. Apr 12, 2009 #2


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    so what is (n+1)!=? (Replace N by n+1)
  4. Apr 12, 2009 #3


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    Note that


    is not correct - you cannot cancel the pair of [tex] 2n [/tex] terms from this. You are
    correct in saying that

    \frac{(n+1)!}{n!} = n+1

    In your second problem, this first step is not correct. That is, you wrote

    \frac{2^{2n+1}}{(2n+1)!} \div \frac{2^{2n-1}}{(2n-1)!} = \frac{2^{2n+1}}{(2n+1)!} \times \frac{2^{2n-1}}{(2n-1)!}

    and this is not correct.
  5. Apr 13, 2009 #4
    Why? So, what do we have to do with it???

    P.S. It was just a typo in my second problem... I figured it out anyway. :smile:

    I understood, thanks!
  6. Apr 13, 2009 #5


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    "Why? So, what do we have to do with it???"

    I'm not sure exactly what you mean - if what you've typed is correct, you can't simplify


    and get [tex] -1 [/tex].
  7. Apr 13, 2009 #6
    If it's wrong then how do you simplify (2n-1)/(2n+1)?

    Yes, what I've typed is correct:


    And the correct answer I should get is:


    If I devide (n+1)!/n! = n+1 which goes on the numerator, what should I do with (2n-1)/(2n+1)?
  8. Apr 13, 2009 #7


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    If it's wrong then how do you simplify (2n-1)/(2n+1)?

    There isn't much you can do - you can't cancel terms that are added, only common factors from a product. The only thing that comes to my mind now is this:

    \frac{2n-1}{2n+1} = \frac{(2n+1)-2}{2n+1} = 1 - \frac{2}{2n+1}

    and that provides you with absolutely no help. This is why I'm confused - I don't know how you are to simplify the expression you are given and obtain your stated answer.

    But again,

    \frac{2n-1}{2n+1} \ne -1

    Think this way: if the left hand side truly did simplify to [tex] -1 [/tex], it would have
    that value for every possible value of [tex] n [/tex] (because it's a constant). But, when
    [tex] n = 10 [/tex], the left hand side is 19/21.

    Perhaps there is a typographical error in your text.
  9. Apr 13, 2009 #8
    Are you sure it's an error??

    here's the actual problem from my textbook:

    http://img23.imageshack.us/img23/635/ddfy.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  10. Apr 13, 2009 #9
    Wait, did you simplify this:
    [tex]\frac{s_{n+1}}{s_{n}} = \frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)}/\frac{n!}{1 \times 3 \times 5 \times ... \times (2n-1)}[/tex]

    to get what you wrote originally in part a), which was this:

    If so, that wasn't simplified correctly. First rewrite as a multiplication:
    [tex]\frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)} \times \frac{1 \times 3 \times 5 \times ... \times (2n-1)}{n!}[/tex]

    1*3*5*...*(2n+1) is the same as 1*3*5*...*(2n-1)*(2n+1), so if you look at the denominator of the 1st fraction and the numerator of the 2nd fraction, everything cancels out except for the 2n+1:
    [tex]\frac{(n+1)!}{2n+1} \times \frac{1}{n!}[/tex]

    You can figure out the rest. (I hope that I didn't give away too much.)

  11. Apr 13, 2009 #10
    Actually I think you meant that it is the same as 1*3*5*...*(2n-1)*(2n)*(2n+1) :biggrin:

    That makes sense now, thank you.


    I have another question (which is very similar to the one above):

    [tex]\frac{(n+1)!}{3\times5 \times7\times...(2n+3)} \times \frac{3\times5\times7\times... \times (2n+1)}{n!}[/tex]

    Does it equal to:


    Since 3*5*7*...*(2n+3) is the same as 3*5*7*...*(2n+1)*(2n+2)*(2n+3)

    Is this right? :rolleyes:
    Last edited: Apr 13, 2009
  12. Apr 13, 2009 #11
    Nope, I meant that 1*3*5*...*(2n+1) = 1*3*5*...*(2n-1)*(2n+1). We are multiplying a series of odd numbers here. 2n-1 is odd, 2n+1 is odd, but 2n is even.

    No. 3*5*7*...*(2n+3) = 3*5*7*...*(2n+1)*(2n+3). 2n+2 shouldn't be there -- it's also even.

  13. Apr 14, 2009 #12
    Oh!! I didn't notice that. So it simplifies to n+1/2n+3. Thanks for your post, it was really helpful!
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