Math Help - sequence

1. 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:

$Xn+1 = Xn \frac{(2n+3)}{(2n+2)}$

2. My attempt

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

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

$x_n \frac{(2n+3)}{(2n+2)} = \frac{8n^3-2n+12n^2-3}{4n^2+4n} \not= X_{n+1}$

3. Hello again,

try to show that $X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$

4. Originally Posted by MPK
Hello again,

try to show that $X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$
Why $Xn$ is $X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$

5. try to prove it

6. Originally Posted by MPK
try to prove it
$Xn$ not is $\frac{(2n-1)(2n+1)}{(2n)} = \frac{4n^2-1}{2n}$ ?

7. No x)

there are many products above and under

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

8. Originally Posted by MPK
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

$X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$

9. try to figure out, you didn't take time

10. Originally Posted by MPK
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

$\frac{(2n-1)!}{(2n)!}$

11. yes it is the idea but you have to calculate more to get a good result,

hint : show that (2n)! = n! * 2^n

12. Originally Posted by MPK
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 $\frac{15}{8}$

13. I find: $\frac{(2n+1)!}{2^n(2n)!}$

14. for n = 2 i have 1/122880

15. Originally Posted by MPK
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 ?
$\frac{(2n+1)!}{2^{2n}(n!)^2} \frac{(2n+3)}{(2n+2)}$

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