Question

prove that (2n)!/n!^2 is an even integer for n>0

Answer 1

Use induction.
First show it's true for n = 1
Then show that if (2n)!/n!^2 is even, then (2(n+1))!/(n+1)!^2 must be even

Answer 2

how about using binomial coefficients?

Answer 3

how u would use induction to show it must be true (2n+2)!/(n+1)!^2

Answer 4

Try to rewrite (2k+2)! and (k+1)! as
blah*blah*(2k)! and (k+1)*k! respectively
I would let you try to figure the blah blah out.

Answer 5

((2k+2)*(2+1) *2k! )/(k+1)*k!

Answer 6

((2k+2)*(2k+1) *2k! )/(k+1)*k!

Answer 7

and what does it give us?

Answer 8

by induct that one part is even.

Answer 9

something*even=still an even number

Answer 10

also you forgot the square part.

Answer 11

\[\frac{(2k+2)(2k+1) \cdot (2k)!}{(k+1)^2 (k!)^2}=\frac{(2k+2)(2k+1)}{(k+1)^2} \cdot \frac{(2k)!}{(k!)^2} \]
But we really didn't even need that induction step. Like we could have looked at that other factor that contains the factor 2, but whateves...either way.

Answer 12

im kind of confused

Answer 13

how would the inductive step look because I'm still confused? So assume that the equation is true then we have to proof that (2(k+1))!/(k+1)!^2 is true?

Answer 14

Yes.
(2(k+1))!=(2k+2)!=(2k+2)*(2k+1)*(2k)!
[(k+1)!]^2=[(k+1)*k!]^2=(k+1)^2*[k!]^2

Answer 15

but then we still don't know that even is multiplied by the integer

Answer 16

You can play with the expression if you want to convince yourself it is an integer.

Answer 17

Re-write the top and the bottom so you see the bottom stuff cancel out.

Answer 18

sorry for so many questions, but i don't see how. we have to leave \[\frac{(2k)!}{(k!)^2}\] alone because thats our induction so we are left with

\[\frac{ (2k+2)(2k+1) }{(k+1)(k+1) }\] and the only way i can think of simplifying that is \[4-\frac{2}{k+1}\] but we are still left with k+1