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=== Fifth proof === This proof is based on the [[Dyck language|Dyck words]] interpretation of the Catalan numbers, so <math>C_n</math> is the number of ways to correctly match {{mvar|n}} pairs of brackets. We denote a (possibly empty) correct string with {{mvar|c}} and its inverse with {{mvar|c'}}. Since any {{mvar|c}} can be uniquely decomposed into <math>c = (c_1) c_2</math>, summing over the possible lengths of <math>c_1</math> immediately gives the recursive definition :<math>C_0 = 1 \quad \text{and} \quad C_{n+1} = \sum_{i=0}^n C_i\,C_{n-i}\quad\text{for }n\ge 0</math>. Let {{mvar|b}} be a balanced string of length {{math|2''n''}}, i.e. {{mvar|b}} contains an equal number of <math>(</math> and <math>)</math>, so <math>\textstyle B_n = {2n\choose n}</math>. A balanced string can also be uniquely decomposed into either <math>(c)b</math> or <math>)c'(b</math>, so :<math>B_{n+1} = 2\sum_{i=0}^n B_i C_{n-i}.</math> Any incorrect (non-Catalan) balanced string starts with <math>c)</math>, and the remaining string has one more <math>(</math> than <math>)</math>, so :<math>B_{n+1} - C_{n+1} = \sum_{i=0}^n {2i+1 \choose i} C_{n-i}</math> Also, from the definitions, we have: :<math>B_{n+1} - C_{n+1} = 2\sum_{i=0}^n B_i C_{n-i} - \sum_{i=0}^n C_i\,C_{n-i} = \sum_{i=0}^n (2B_i-C_i) C_{n-i}.</math> Therefore, as this is true for all {{mvar|n}}, :<math>2B_i - C_i = \binom{2i+1}{i}</math> :<math>C_i = 2B_i - \binom{2i+1}{i}</math> :<math>C_i = 2\binom{2i}{i} - \binom{2i+1}{i}</math> :<math>C_i=\frac{1}{i+1}\binom{2i}{i}</math>
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