Alternating Sum and Difference of Binomial Coefficients for Given n

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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.


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

$\blacksquare$


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