Catalan number proof

Theorem: 2n a1, a2… which consists of n-1 and n-1 .... A2n, its part and

 

Meet a1 + a2 +... AK> = 0 (k = 1 2... 2n) the number of conditions is

 

 

 

Proof: Make An acceptable series Bn unacceptable Series

 

So An + Bn =. An is what we need. We can obtain An by finding Bn.

 

Let's assume that there is a minimum K that makes a1 + a2 +... + If ak is negative, you can be sure that a1 +... + Ak-1 = 0, and ak =-1.k is an odd integer.

 

For each nonconforming sequence a1 + a2 +... + Ak +... + A2n replace a1 a2 ak with-a1-a2-ak, then the new series

 

A1 'a2 '... A2n has n + 1 + 1 and n-1-1. That is, each non-conforming series will change

 

N + 1 + 1 and n-1-1. Now we need to prove the number of n + 1 + 1 and n-1-1 = Bn

 

The arrangement of n + 1 + 1 and n-1-1 must have a minimum k so that a1 + a2 +... AK <0 and this part is also replaced by-a1-a2-ak, which becomes the 2n item a1, a2… composed of n 1 and n-1 .... A2n and the arrangement of Bn is easy to find.

 

 

 

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Application: There are 2n people who want to go to the theater, 5 corners for admission, 1 RMB for n people, 5 corners for n people, and no change in the ticketing office at the beginning of the theater, the number of queue methods.

 

I feel a little far-fetched in bold proof. please correct me if you have any good methods.