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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 !

D S
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An_Elephant
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    well, your results are wrong. For example, let $S_n=\sum_{k=0}^n(-1)^k\binom{n}{k}$. Then $S_0=1$, $S_2=1$... – Surb Apr 28 '23 at 09:10
  • Sorry, I forgot to mention that $n>2$ . – An_Elephant Apr 28 '23 at 09:13
  • Maybe you can try to differentiate $(x-1)^n$ using the product rule on the one hand and differentiating its binomial expansion on the other side, then set $x=1.$ – F_M_ Apr 28 '23 at 09:32
  • @F_M_ But differentiation or integration will make it messy because extraneous coefficients will come. – An_Elephant Apr 28 '23 at 10:27
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    https://math.stackexchange.com/questions/3833276/if-px-is-any-polynomial-of-degree-less-than-n-show-that-sum-j-0n-1/3833675#3833675 – RobPratt Apr 28 '23 at 14:27
  • @RobPratt Thanks ! It was helpful. – An_Elephant Apr 28 '23 at 17:07
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    The answers in the duplicate question prove that $S(a)=(-1)^n {a\brace n}n!$, where ${a\brace n}$ is a Stirling number of the second kind, equal to the number of ways to partition $a$ distinct items into $n$ unlabeled, nonempty groups. It is obvious by this combinatorial definition that ${a\brace n}$ is zero when $a<n$ (cannot put $4$ people in $5$ nonempty groups), and that ${a\brace n}$ is nonzero when $a\ge n$. – Mike Earnest Apr 28 '23 at 17:09

1 Answers1

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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 $$\sum_{k=0}^n (-1)^k k^n \binom{n}{k} = (-1)^n n! \ne 0.$$

Sean Eberhard
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  • It went above my head :( Anyway thanks, I hope I will understand your solution in future when I will learn more advanced mathematics. – An_Elephant Apr 28 '23 at 10:20
  • Hi, Is it true that $S$ is strictly non-zero whenever $m>n$ ? Thanks ! – An_Elephant Apr 28 '23 at 17:05
  • @An_Elephant Yes, that's true, though it's not quite clear from my answer. See the answers to the duplicate question https://math.stackexchange.com/questions/1875714/simplifying-sum-r-0n-n-chooserrk-1r – Sean Eberhard Apr 30 '23 at 09:30
  • Okay thanks. I get it directly from combinatorial proof. – An_Elephant Apr 30 '23 at 12:39