Increasing Alternating Sum of Binomial Coefficients

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Theorem

Let $n \in \Z$ be an integer.


Then:

$\ds \sum_{j \mathop = 0}^n \paren {-1}^{n + 1} j \binom n j = 0$

where $\dbinom n k$ denotes a binomial coefficient.


That is:

$1 \dbinom n 1 - 2 \dbinom n 2 + 3 \dbinom n 3 - \cdots + \paren {-1}^{n + 1} n \dbinom n n = 0$


Proof

\(\ds \sum_{j \mathop = 0}^n \paren {-1}^{n + 1} j \binom n j\) \(=\) \(\ds \sum_{j \mathop = 1}^n \paren {-1}^{n + 1} j \binom n j\) as $0 \dbinom n 0 = 0$
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^n \paren {-1}^{n + 1} n \binom {n - 1} {j - 1}\) Factors of Binomial Coefficient
\(\ds \) \(=\) \(\ds n \sum_{j \mathop = 0}^{n - 1} \paren {-1}^{n - 1} \binom {n - 1} j\) Translation of Index Variable of Summation, and $\paren {-1}^{n + 1} = \paren {-1}^{n - 1}$
\(\ds \) \(=\) \(\ds 0\) Alternating Sum and Difference of Binomial Coefficients for Given n

$\blacksquare$


Sources

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