Question

Prove that \(\forall n \in \mathbb{N}, n \geq 1\)
\( \Large \prod_{k=1}^n (2k-1) = \frac{(2n)!}{2^{n}.n!}\)

I need just solution for n->n+1

Answer 1

so ∏ n k=1 (2k−1)=(2n)! 2 n .n!
for some n in N

Answer 2

so we have (2(n+1)-1)*((2n)!)/(2^n*n!) <-- you understnad how i got here for n+1? its our inductive hypothisis and the fact that ∏ from k=1 to n+1 is the same thing as (2(n+1)-1)*∏ from k = 1 to n

Answer 3

yes i understand i done same thing as you, what would be the next step ?

Answer 4

(2(n+1)-1)*((2n)!)/(2^n*n!) = (2n+1)(2n)!/(2^n*n!)
= (2n+2)(2n+1)(2n!)/(2^n*(2n+2)*n!) = (2n+2)!/(2^n*2*(n+1)*(n!))
= (2(n+1))!/(2^(n+1)*(n+1)!)

Answer 5

make sense? I got to go to bed soon

Answer 6

i have problem after second equal sign

Answer 7

just multiplied both top and bottom by (2n+2)

Answer 8

I did not just know to do that, I worked the RHS on paper and then showed you the backwards steps to get there.

Answer 9

its ok to do this step because 2n+2 will only be 0 when n = -1 and n is in N so this will never happen

Answer 10

hmm i think its right solution but i need some other way, i had seen the solution of this question by some friend and the end solution was shorter then this one and clearer for me, but thanks for your effort

Answer 11

if you stay on the lhs you can get to (2n+1)(2n)!/(2^n*n!)

and if you start on the RHS you can get to (2n+2)!/(2^n*2*(n+1)*(n!)) and then (2n+1)(2n)!/(2^n*n!)

try to do this on paper and see how it goes from there.

Answer 12

good luck man, I think I could write this so its nicer but it would take me an hour with the editor:)

Answer 13

no problem thank you zzrOrck3r :)

Answer 14

hope that helps...

Answer 15

thanks that looks much better ;)

Answer 16

lol:) aight gnight

Answer 17

thank you if you didnt slept already i have one question :)
how this possible?
\[ \Large 2^{n+1}.(n+1)! = 2^{n}.2.(n+1)(n!)\]

Answer 18

for my logic in right side \[ \Large (n!)\] is too much

Answer 19

2^(n+1) = 2^n * 2 you cool with that part?

Answer 20

yes

Answer 21

(n+1)! = (n+1)*(n+1-1)! = (n+1)*(n)! just like
3! = 3*(3-1)! = 3 * 2!

Answer 22

(n+3)! = (n+3)(n+2)(n+1)n!

Answer 23

ok zzrOck3r i think you are right, my head is too thick to get sometimes such things ;)

Answer 24

its cool we all do that the first time we encounter that issue: but yes its the same thing as
5! = 5*4*3!
aight you will be fine:)

Answer 25

thx :)

Answer 26

np

Answer 27

Here is a proof without induction
\[
(2n)! = 1\, 2\, 3\, 4\, 5 \cdots (2n -1)\, 2n=
\\
1\, 3\, , 5 \cdots (2n -1)\, 2\, 4\, \cdots \, 2n=
\\
\prod_{k=1}^{n} (2k-1)\, (2)\,1\, (2)\,2\, (2)\,3 \cdots \,(2)n=\\
\prod_{k=1}^{n} (2k-1)\, 2^n n!\\
\\
\frac{(2n)!}{2^n n!}=\prod_{k=1}^{n} (2k-1)

\]

Answer 28

thank you mr Saab, your answer helped

Answer 29

yw