Alternating Sum and Difference of Binomial Coefficients for Given n

Theorem

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{i=0}^n \left({-1}\right)^i \binom n i\)	\(=\)	\(\displaystyle \binom n 0 + \sum_{i=1}^{n-1} \left({-1}\right)^i \binom n i + \left({-1}\right)^n \binom n n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom n 0 + \sum_{i=1}^{n-1} \left({-1}\right)^i \left({\binom {n-1} {i-1} + \binom {n-1} i}\right) + \left({-1}\right)^n \binom n n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Pascal's Rule
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom n 0 - \binom {n-1} 0 + \left({-1}\right)^{n-1} \binom {n-1} {n-1} + \left({-1}\right)^n \binom n n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	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:

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

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

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

$\blacksquare$

Proof 3

The assertion can be expressed:

$\displaystyle \sum_{i \le n} \left({-1}\right)^i \binom n i = 0$ 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:

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

Putting $r = n$ we have:

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

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

$\blacksquare$

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.

$\blacksquare$

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $3.8$