How to prove this indentity $\binom{100}{0}^2-\binom{100}{1}^2+\binom{100}{2}^2-...-\binom{100}{99}^2+\binom{100}{100}^2=\binom{100}{50}$

Asked Apr 10 '14 at 08:16

Active Apr 10 '14 at 10:07

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I don't know how to prove this identity: $\binom{100}{0}^2-\binom{100}{1}^2+\binom{100}{2}^2-\binom{100}{3}^2+...-\binom{100}{99}^2+\binom{100}{100}^2=\binom{100}{50}$

edited Apr 10 '14 at 10:07

Martin Sleziak

53,687

asked Apr 10 '14 at 08:16

mathlover

Are you sure that this identity even holds? – 5xum Apr 10 '14 at 08:27
@5xum A simple PARI-GP script verifies that the above identity holds true. – Yiyuan Lee Apr 10 '14 at 08:40
Yes, I tried it myself. First I tried to replace $100$ with $2$ and id didn't hold, so I got a bit worried. I now agree. – 5xum Apr 10 '14 at 08:41
Apparently this already has a detailed answer here. – Start wearing purple Apr 10 '14 at 08:47

How to prove this indentity $\binom{100}{0}^2-\binom{100}{1}^2+\binom{100}{2}^2-...-\binom{100}{99}^2+\binom{100}{100}^2=\binom{100}{50}$

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