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: $\displaystyle Xn+1 = Xn \frac{(2n+3)}{(2n+2)}$
Follow Math Help Forum on Facebook and Google+
My attempt $\displaystyle X_n = \frac{4n^2-1}{2n}$ $\displaystyle X_{n+1} = \frac{(2n+2-1)(2n+2+1)}{2n+2} = \frac{4n^2+8n+3}{2n+2}$ $\displaystyle x_n \frac{(2n+3)}{(2n+2)} = \frac{8n^3-2n+12n^2-3}{4n^2+4n} \not= X_{n+1}$
Hello again, try to show that $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$
Originally Posted by MPK Hello again, try to show that $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$ Why $\displaystyle Xn$ is $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$
try to prove it
Originally Posted by MPK try to prove it $\displaystyle Xn$ not is $\displaystyle \frac{(2n-1)(2n+1)}{(2n)} = \frac{4n^2-1}{2n}$ ?
No x) there are many products above and under you have 1*3*5*7***(2n-1)*(2n+1) and etc
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 $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$
try to figure out, you didn't take time
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 $\displaystyle \frac{(2n-1)!}{(2n)!}$
yes it is the idea but you have to calculate more to get a good result, hint : show that (2n)! = n! * 2^n
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 $\displaystyle X_n = \frac{(2n+1)!}{2^{2n}(n!)^2}$ if I use $\displaystyle n=2$ I have $\displaystyle \frac{15}{8}$
I find: $\displaystyle \frac{(2n+1)!}{2^n(2n)!}$
for n = 2 i have 1/122880
Originally Posted by MPK 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 ? $\displaystyle \frac{(2n+1)!}{2^{2n}(n!)^2} \frac{(2n+3)}{(2n+2)}$
View Tag Cloud