Question

help with these two problems - binomial thm
http://prntscr.com/3io3dg

Answer 1

@BSwan

Answer 2

\[\binom{2n}{n} = \dfrac{(2n)!}{n! n!} = \dfrac{[1.3.5...(2n-1)][2.3.6...(2n)]}{n!n!}\]

Answer 3

\[ = \dfrac{[1.3.5...(2n-1)][2^n(1.2.3...(n)]}{n!n!} \\= \dfrac{[1.3.5...(2n-1)][2^nn!]}{n!n!} \\ = \dfrac{[1.3.5...(2n-1)][2^n]}{n!} \]

Answer 4

\[ = \dfrac{[1.3.5...(2n-1)][2^n]}{1.2.3...n} \]

\[ = \dfrac{[1.3.5...(2n-1)][2^{2n}]}{2.4.6...2n} \]

Answer 5

did u get the hint in #9 ?

Answer 6

im trying to match u with 8 huh !
ok ill move on 9

Answer 7

ohkay, i made a typo in the first step sorry. corrected below :

\[\binom{2n}{n} = \dfrac{(2n)!}{n! n!} = \dfrac{[1.3.5...(2n-1)][2.4.6...(2n)]}{n!n!}\]

Answer 8

^^ ive just separated the "odd" and "even" prducts...

Answer 9

ok i got 9

Answer 10

cool, im still trying to figure out what the hint is about

Answer 11

well its crap lol

Answer 12

oh, got u... that makes sense now xD

Answer 13

haha oki

Answer 14

so the only step we need to check is