# 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

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

## Proof 1

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \sum_{i \mathop = 0}^n \left({-1}\right)^i \binom n i$$ $$=$$ $$\displaystyle$$ $$\displaystyle \binom n 0 + \sum_{i \mathop = 1}^{n-1} \left({-1}\right)^i \binom n i + \left({-1}\right)^n \binom n n$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \binom n 0 + \sum_{i \mathop = 1}^{n-1} \left({-1}\right)^i \left({\binom {n-1} {i-1} + \binom {n-1} i}\right) + \left({-1}\right)^n \binom n n$$ $$\displaystyle$$ $$\displaystyle$$ Pascal's Rule $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \binom n 0 - \binom {n-1} 0 + \left({-1}\right)^{n-1} \binom {n-1} {n-1} + \left({-1}\right)^n \binom n n$$ $$\displaystyle$$ $$\displaystyle$$ as the series telescopes

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.

$\blacksquare$

## Proof 2

From the Binomial Theorem, we have that:

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

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

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

$\blacksquare$

## 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.

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

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 $\displaystyle \binom {n - 1} n = 0$.

$\blacksquare$