Prove identity based on binomial theorem

Question

$\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.

http://math.stackexchange.com/questions/941857/problem-proving-sum-r-0n-1-binom2n-1r-22n-2 — lab bhattacharjee, Sep 27 '14 at 06:06

score 1 · Accepted Answer · answered Sep 27 '14 at 05:47

1

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

answered Sep 27 '14 at 05:47

Yuval Filmus

But how to relate the 2n-1 and the n-1? – Wang Kah Lun Sep 27 '14 at 05:52
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I gave you a hint, now it's your turn to continue thinking. – Yuval Filmus Sep 27 '14 at 05:52

1 Answers1