Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - sequence

  1. #1
    Super Member
    Joined
    Jun 2008
    Posts
    829

    sequence

    Consider the sequence of real numbers defined by:
    Xn = \frac{1.3.5.7...(2n-1)(2n+1)}{2.4.6...(2n)}, n \in N , n \not= 0


    Prove that:

    Xn+1 = Xn \frac{(2n+3)}{(2n+2)}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    829
    My attempt

    X_n = \frac{4n^2-1}{2n}

    X_{n+1} = \frac{(2n+2-1)(2n+2+1)}{2n+2} = \frac{4n^2+8n+3}{2n+2}

    x_n \frac{(2n+3)}{(2n+2)} = \frac{8n^3-2n+12n^2-3}{4n^2+4n} \not= X_{n+1}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MPK
    MPK is offline
    Newbie
    Joined
    Sep 2009
    Posts
    10
    Hello again,

    try to show that X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jun 2008
    Posts
    829
    Quote Originally Posted by MPK View Post
    Hello again,

    try to show that X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}
    Why Xn is X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MPK
    MPK is offline
    Newbie
    Joined
    Sep 2009
    Posts
    10
    try to prove it
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2008
    Posts
    829
    Quote Originally Posted by MPK View Post
    try to prove it
    Xn not is \frac{(2n-1)(2n+1)}{(2n)} = \frac{4n^2-1}{2n} ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MPK
    MPK is offline
    Newbie
    Joined
    Sep 2009
    Posts
    10
    No x)

    there are many products above and under

    you have 1*3*5*7***(2n-1)*(2n+1) and etc
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Jun 2008
    Posts
    829
    Quote Originally Posted by MPK View Post
    No x)

    there are many products above and under

    you have 1*3*5*7***(2n-1)*(2n+1) and etc
    Oh yes. But I do not understand how you found

    X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MPK
    MPK is offline
    Newbie
    Joined
    Sep 2009
    Posts
    10
    try to figure out, you didn't take time
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Jun 2008
    Posts
    829
    Quote Originally Posted by MPK View Post
    try to figure out, you didn't take time
    I think:

    The numerator => multiplication between the odd numbers
    The denominator => multiplication between the numbers pairs

    \frac{(2n-1)!}{(2n)!}
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MPK
    MPK is offline
    Newbie
    Joined
    Sep 2009
    Posts
    10
    yes it is the idea but you have to calculate more to get a good result,


    hint : show that (2n)! = n! * 2^n
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Jun 2008
    Posts
    829
    Quote Originally Posted by MPK View Post
    yes it is the idea but you have to calculate more to get a good result,


    hint : show that (2n)! = n! * 2^n
    OK. In this equation X_n = \frac{(2n+1)!}{2^{2n}(n!)^2} if I use n=2 I have \frac{15}{8}
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Jun 2008
    Posts
    829
    I find: \frac{(2n+1)!}{2^n(2n)!}
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MPK
    MPK is offline
    Newbie
    Joined
    Sep 2009
    Posts
    10
    for n = 2 i have 1/122880
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Super Member
    Joined
    Jun 2008
    Posts
    829
    Quote Originally Posted by MPK View Post
    for n = 2 i have 1/122880

    X_{n+1} = \frac{(2n+3)!}{2^{2n+2}((n+1)!)^2}

    Problem:
    1) Prove that X_{n+1} = X_n \frac{(2n+3)}{(2n+2)}

    How can I multiply factorial ?
     \frac{(2n+1)!}{2^{2n}(n!)^2} \frac{(2n+3)}{(2n+2)}
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 24th 2010, 02:10 AM
  2. Replies: 0
    Last Post: July 4th 2010, 12:05 PM
  3. Replies: 2
    Last Post: March 1st 2010, 11:57 AM
  4. sequence membership and sequence builder operators
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: June 4th 2009, 03:16 AM
  5. Replies: 12
    Last Post: November 15th 2006, 12:51 PM

Search Tags


/mathhelpforum @mathhelpforum