Alternating Sum and Difference of Binomial Coefficients for Given n

Theorem

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

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

Proof 1
$$ $$ $$

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

Hence the result.

Proof 2
From the Binomial Theorem, we have that:


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

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

$$ $$ $$

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

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

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

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

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