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$\displaystyle \sum_{r=0}^{n-1} {2n-1 \choose r} = 2^{2n-2} $

Perhaps it can be proved by using sum of all combinations from r=0 to r=n is 2 to the power of n.

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Hint: $$ 2\sum_{r=0}^{n-1} \binom{2n-1}{r} = \sum_{r=0}^{n-1} \left[\binom{2n-1}{r} + \binom{2n-1}{2n-1-r}\right]. $$

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  • $\begingroup$ But how to relate the 2n-1 and the n-1? $\endgroup$ – Alan Wang Sep 27 '14 at 5:52
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    $\begingroup$ I gave you a hint, now it's your turn to continue thinking. $\endgroup$ – Yuval Filmus Sep 27 '14 at 5:52

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