Background:
Let $n>2$ and $ n \in N$.
$$\sum_{k = 0}^n (-1)^k \binom nk (x-k)(y-k)(z-k)= 0$$ (not verbatim)
I was solving the above textbook problem. As a part of my solution, I came across the below sums , which I evaluated successfully , used them back in my solution and my final answer also matched with the answer key :
$$\sum_{k=0}^n (-1)^k \binom nk = 0$$ $$\sum_{k=0}^n (-1)^k k \binom nk = 0 $$ $$\sum_{k=0}^n (-1)^k k^2\binom nk = 0$$ $$\sum_{k=0}^n (-1)^k k^3\binom nk = 0$$
Even though I successfully solved the textbook problem, but I noticed a pattern in this part of my solution. The above sums are all zero and of the form :
$$S(a) = \sum_{k=0} ^ n(-1)^k k^a \binom nk$$
Question:
As said, $S(0), S(1) , S(2)$ and $S(3)$ are all zero. I want to know if $S(a)$ is always zero $\forall a \in N$ ? If No, then are there any special pattern of natural numbers for which the sum is zero ?
Thanks !
One can show that $$\sum_{k=0}^n (-1)^k k^m \binom{n}{k} = 0$$ for $n > m$. Equivalently, $$\sum_{k=0}^n (-1)^k p(k) \binom{n}{k} = 0$$ for any polynomial $p$ of degree $\deg p < n$. Since the space of polynomials of degree $< n$ is spanned by $\binom{k}{0}, \dots, \binom{k}{n-1}$, it suffices to prove that $$\sum_{k=0}^n (-1)^k \binom{k}{m} \binom{n}{k} = 0$$ for $m < n$. Since $\binom{k}{m} \binom{n}{k} = \binom{n}{m} \binom{n-m}{k-m}$ we indeed have $$\sum_{k=0}^n (-1)^k \binom{k}{m} \binom{n}{k} = \binom{n}{m} \sum_{k=m}^n (-1)^k \binom{n-m}{k-m} = 0.$$
On the other hand for $m=n$ we have $$\sum_{k=0}^n (-1)^k \binom{k}{n} \binom{n}{k} = (-1)^n,$$ so $$\sum_{k=0}^n (-1)^k k^n \binom{n}{k} = (-1)^n n! \ne 0.$$