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I am struggling to verify the following identity.

$$\binom{2m}{m} \frac{m}{2} = \sum_{j=1}^m j \binom{2m}{m+j}$$

I've tried induction, but I run into issues inside the sum. I can't see a combinatorial interpretation either. I've noted the right-hand side can be rewritten as $\frac{1}{2}\sum_{j=-m}^m |j| \binom{2m}{m+j}$, but this seems more complicated. Any hints would be appreciated!

Aside: my goal is actually to compute the equality in this answer (expected distance of a one-dimensional random walk). Is there a more direct way to verify this?

angryavian
  • 89,882

1 Answers1

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We use \begin{align*} \sum_{j=0}^{m-1} \binom{2m}{j} & = \sum_{j=m+1}^{2m} \binom{2m}{j} = \frac{1}{2} \left( 2^{2m} - \binom{2m}{m} \right) \\ % \sum_{j=0}^{m-1} \binom{2m-1}{j} & = \sum_{j=m}^{2m-1} \binom{2m-1}{j} = \frac{1}{2} \cdot 2^{2m-1}. \end{align*}

Then

\begin{align*} \sum_{j=1}^m j \binom{2m}{m+j} & = \sum_{j=1}^m (m+j) \binom{2m}{m+j} - \sum_{j=1}^m m \binom{2m}{m+j} \\ % & = 2m \sum_{j=0}^{m-1} \binom{2m-1}{m+j} - m \left( 2^{2m} - \binom{2m}{m} \right) \cdot \frac{1}{2} \\ % & = 2m \cdot 2^{2m-1}\cdot \frac{1}{2} - m \cdot 2^{2m-1}+ \frac{m}{2} \binom{2m}{m} \\ % & = \frac{m}{2}\binom{2m}{m}. % \end{align*}

Catalin Zara
  • 6,187