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Thread: sequence

  1. #1
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    sequence

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


    Prove that:

    $\displaystyle Xn+1 = Xn \frac{(2n+3)}{(2n+2)}$
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  2. #2
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    My attempt

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

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

    $\displaystyle x_n \frac{(2n+3)}{(2n+2)} = \frac{8n^3-2n+12n^2-3}{4n^2+4n} \not= X_{n+1}$
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  3. #3
    MPK
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    Hello again,

    try to show that $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$
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  4. #4
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    Quote Originally Posted by MPK View Post
    Hello again,

    try to show that $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$
    Why $\displaystyle Xn$ is $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$
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  5. #5
    MPK
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    try to prove it
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  6. #6
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    Quote Originally Posted by MPK View Post
    try to prove it
    $\displaystyle Xn$ not is $\displaystyle \frac{(2n-1)(2n+1)}{(2n)} = \frac{4n^2-1}{2n}$ ?
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  7. #7
    MPK
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    No x)

    there are many products above and under

    you have 1*3*5*7***(2n-1)*(2n+1) and etc
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  8. #8
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    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

    $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$
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  9. #9
    MPK
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    try to figure out, you didn't take time
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  10. #10
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    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

    $\displaystyle \frac{(2n-1)!}{(2n)!}$
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  11. #11
    MPK
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    yes it is the idea but you have to calculate more to get a good result,


    hint : show that (2n)! = n! * 2^n
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  12. #12
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    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 $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$ if I use $\displaystyle n=2$ I have $\displaystyle \frac{15}{8}$
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  13. #13
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    I find: $\displaystyle \frac{(2n+1)!}{2^n(2n)!}$
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  14. #14
    MPK
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    for n = 2 i have 1/122880
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  15. #15
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    Quote Originally Posted by MPK View Post
    for n = 2 i have 1/122880

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

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

    How can I multiply factorial ?
    $\displaystyle \frac{(2n+1)!}{2^{2n}(n!)^2} \frac{(2n+3)}{(2n+2)}$
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