Is there a function $p:\mathbb N\to \{ 1,-1 \} $ and a fixed $N\in \mathbb N$ such that for every $n \geq N$ we get:


$\sum \_{i=0} ^{n} p(i)\binom {n}{i}=0$
?


Obviously $p(i)=(-1)^i$ works for $N=1$, and so does $p(i)=(-1)^{i+1}$, but are there any others?


(my personal guess is that there aren't)



Zero sum of binomial coefficients
No, there are no others.


In fact, define a function $q : \mathbb N\to\left\lbrace 1,-1\right\rbrace$ by $q\left(i\right) = \left(-1\right)^i p\left(i\right)$ for every $i\in\mathbb N$. Then, $\sum\limits\_{i=0}^n p\left(i\right) \binom{n}{i} = 0$ becomes $\sum\limits\_{i=0}^n \left(-1\right)^i q\left(i\right) \binom{n}{i} = 0$. Now, denote, for every $n,x\in\mathbb N$, the number $\sum\limits\_{i=0}^n \left(-1\right)^i q\left(x+i\right) \binom{n}{i}$ by $\left(\Delta^n q\right)\left(x\right)$. Then, the function $\Delta^n q$ thus defined is the so-called *$n$-th finite difference* of the function $q$. (If you don't know about finite differences, this is the right time to google.) Our condition $\sum\limits\_{i=0}^n \left(-1\right)^i q\left(i\right) \binom{n}{i} = 0$ rewrites as $\left(\Delta^n q\right)\left(0\right) = 0$.


So we have $\left(\Delta^n q\right)\left(0\right) = 0$ for all $n\geq N$. We will now prove that $\Delta^n q = 0$ (not only at the point $0$, but as functions!) for all $n\geq N$. In fact, let us prove that every $x\in \mathbb N$ satisfies $\left(\Delta^n q\right)\left(x\right) = 0$ for all $n\geq N$. We will prove this by induction over $x$: The induction base ($x=0$) follows from $\left(\Delta^n q\right)\left(0\right) = 0$. For the induction step, assume that some $x\in\mathbb N$ satisfies $\left(\Delta^n q\right)\left(x\right) = 0$ for all $n\geq N$. Then, clearly, $\left(\Delta^{n+1} q\right)\left(x\right) = 0$ for all $n\geq N$ as well (because $n\geq N$ entails $n+1\geq N$). Now, the properties of finite differences yield


$\left(\Delta^{n+1} q\right)\left(x\right) = \left(\Delta\left(\Delta^n q\right)\right)\left(x\right) = \left(\Delta^n q\right)\left(x\right) - \left(\Delta^n q\right)\left(x+1\right)$.


Since $\left(\Delta^{n+1} q\right)\left(x\right) = 0$ and $\left(\Delta^n q\right)\left(x\right) = 0$, we thus obtain $\left(\Delta^n q\right)\left(x+1\right) = 0$. This completes the induction step. Hence, we have shown that every $x\in \mathbb N$ satisfies $\left(\Delta^n q\right)\left(x\right) = 0$ for all $n\geq N$. Thus, $\Delta^n q = 0$ for all $n\geq N$. In particular, $\Delta^N q=0$. Now, by another well-known property of finite differences, a function whose $N$-th finite difference is $0$ must be a polynomial of degree less than $N$. Hence, $q$ is a polynomial of degree less than $N$. Since $q$ only takes values in $\left\lbrace 1,-1\right\rbrace$, this yields that $q$ is either constantly $1$ or constantly $-1$. This means that $p$ is either $\left(-1\right)^i$ or $\left(-1\right)^{i+1}$.


