Question

Can you calculate this to the next step ?
\[\frac{(2n+2)(2n+1)(2n)!}{(2n+2)2^{n}n!} = ? \]

Answer 1

\[\frac{(2n+2)(2n+1)(2n)!}{(2n+2)2^{n}n!} =\frac{(2n+1)(2n)!}{2\times 4\times 6\times...\times2n} \]make sense till here?

Answer 2

2x3x6x... this part not..

Answer 3

actually \[(2n+2)2^{n}n! = ?\] would help me too..

Answer 4

\[\ \ \ n!=1\times2\times3\times...\times n \]\[\ \ \ 2^n=2\times2\times2\times...\times 2 \]\[2^n n!=2\times4\times6\times...\times2n\]

Answer 5

ok..

Answer 6

\[\frac{(2n+2)(2n+1)(2n)!}{(2n+2)2^{n}n!} =\frac{(2n+1)(2n)!}{2\times 4\times 6\times...\times2n}=(2n+1)\times(2n-1)\times...\times5\times3\times1\]

Answer 7

is it not possible to show it like that for example
\[(2n+2)2^{n}n! = 4n^{n} + 4^{n} (n!)\]
and i dont know if its right.. ?

Answer 8

its impossible i think

Answer 9

ok thanks anyway

Answer 10

welcome :)