Alternating Sum and Difference of Binomial Coefficients for Given n

Theorem

 * $\displaystyle \sum_{i \mathop = 0}^n \left({-1}\right)^i \binom n i = 0$ for all $n \in \Z: n > 0$

where $\displaystyle \binom n i$ is a binomial coefficient.

Corollary

 * $\displaystyle \sum_{i \in \Z} \left({-1}\right)^i \binom n i = 0$

Proof 1
We note:
 * $\displaystyle \binom n 0 = \binom {n-1} 0 = 1$ so $\displaystyle \binom n 0 - \binom {n-1} 0 = 0$;
 * $\displaystyle \left({-1}\right)^{n-1} \binom {n-1} {n-1} = - \left({-1}\right)^n \binom n n = - \left({-1}\right)^n$ so $\displaystyle \left({-1}\right)^{n-1} \binom {n-1} {n-1} + \left({-1}\right)^n \binom n n = 0$.

Hence the result.

Proof 2
From the Binomial Theorem, we have that:


 * $\displaystyle \forall n \in \Z_+: \left({x+y}\right)^n = \sum_{i \mathop = 0}^n \binom n i x^{n-i} y^i$

Putting $x = 1, y = -1$, we get:

Proof 3
The assertion can be expressed:
 * $\displaystyle \sum_{i \mathop \le n} \left({-1}\right)^i \binom n i = 0$ for all $n > 0$

as $\displaystyle \binom n i = 0$ when $i < 0$ by definition of binomial coefficient.

From Alternating Sum and Difference of r Choose k up to n we have:
 * $\displaystyle \sum_{i \mathop \le n} \left({-1}\right)^i \binom r i = \left({-1}\right)^n \binom {r - 1} n$

Putting $r = n$ we have:
 * $\displaystyle \sum_{i \mathop \le n} \left({-1}\right)^i \binom n i = \left({-1}\right)^n \binom {n - 1} n$

As $n-1 > n$ it follows from the definition of binomial coefficient that $\displaystyle \binom {n - 1} n = 0$.

Proof of Corollary
From the definition of the binomial coefficient, when $i < 0$ and $i > n$ we have $\displaystyle \binom n i = 0$.

The result follows directly.