Originally Posted by

**blaise** This was the best fit for Binomial Theorem I could find--Probability & Stats, hopefully it's in the right spot.

The question:

Use the binomial expansion of (a + b)^n and the identity (1 - 1)^n = 0 to prove

(n choose 0) - (n choose 1) + (n choose 2) - ... + (-1)^2(n choose n) = 0, where n != 0

I perfectly understand the varying signs and why it works out like that in my head but I can't prove it. My proving sucks bad. I think in the obvious common sense way that's usually wrong because I assume a buncha things.

I'm thinking:

(n choose 0) - (n choose 1) + (n choose 2) - ... + (-1)^2(n choose n) = 0

R.S.

Multiply it by a.

a(n choose 0) - a(n choose 1) + a(n choose 2) - ... + a(-1)^2(n choose n)

Through inspection

(a - a)^n

(0)^n

0

L.S. = R.S.

0 = 0

But I'm pretty sure that's wrong... Any ideas/suggestions =)

And expect more from me in the near future with random pesterings of more proving questions cause I'm super hopeless at those.

Thanks a lot in advance.