# Thread: Proving Binomial Theorem (1-1)^n = 0

1. ## Proving Binomial Theorem (1-1)^n = 0

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.

2. 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.

$(1-x)^n=\sum_{k=0}^n {n \choose k}(-x)^k$
So when $x=1$, we have:
$(1-1)^n=\sum_{k=0}^n {n \choose k}(-1)^k=0$