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How can I prove the identity $\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}$?

I have to prove it using lattice paths, it should be related to Catalan numbers

The $n$th Catalan number $C_n$ counts the number of monotonic paths along the edges of a grid with $n\times n$ square cells, which do not pass above the diagonal.

See for example this link

For example $\frac{1}{k}\binom{2k-2}{k-1}$ is exactly $C_{k-1}$, and the other terms can also be expressed in terms of the Catalan numbers.

The second part of the exercise ask to prove the recurrence formula $C_n=\sum_{k=1}^n C_{k-1}C_{n-k}$ using similar reasoning (i.e. lattice paths). So we can't use this formula to prove the first.

Could you help me please?

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The right hand side counts the number of paths from $(0,0)$ to $(n+1,n-1)$, so I suspect the lhs does too, but in a different way. –  TMM Sep 20 '11 at 1:46
$\dfrac{1}{k}\dbinom{2k-2}{k-1}$ is actually $C_{k-1}$. –  Brian M. Scott Sep 20 '11 at 1:53
@Brian: yeah, you're right, I will fix the question –  Alex M Sep 20 '11 at 2:34
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1 Answer

up vote 7 down vote accepted

The right-hand side counts the number of monotonic paths from $(0,0)$ to $(n-1,n+1)$. Since $(n-1,n+1)$ is above the diagonal, every one of these paths must cross the diagonal at some point. Suppose that the first ‘bad’ step is from $(k-1,k-1)$ to $(k-1,k)$.

  • How many ways are there to get from $(0,0)$ to $(k-1,k-1)$ without going above the diagonal?

  • How many ways are there to get from $(k-1,k)$ to $(n-1,n+1)$?

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