# 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.

Thanks a lot in advance.

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.

Thanks a lot in advance.
By the binomial theorem:

$(1-x)^n=\sum_{k=0}^n {n \choose k}(-x)^k$

So when $x=1$, we have:

$(1-1)^n=\sum_{k=0}^n {n \choose k}(-1)^k=0$

which is the required result.

RonL