Let $m \in\mathbb N$. Define a polynom $P$ by: $$ P(x)=\sum_{k=0}^{m+1} \binom{m+1}{k}(-1)^k (x-k)^m $$ Prove that $P(x)\equiv 0$. I tryed to use taylor polynomials, finding roots, but it did not help. Any suggestions?
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Expand $(x-k)^m$ by the binomial theorem as $\sum_{r=0}^mx^r(-k)^{m-r}{m\choose r}$. Then look at the terms in $x^r$ in the double sum. Ignoring the factors which do not involve $r$ they are simply $\sum_{k=0}^{m+1}{m+1\choose k}(-1)^k$ which is 0 by the binomial theorem.
almagest
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Induction on $m$. $m=0$ is obvious
Assume m-1 is ok.
Let's prove for m
${m+1 \choose k} = {m \choose k} + {m \choose k-1} $
$ \sum_{k=0}^{m+1} {m+1 \choose k} (-1)^k (x-k)^m = \sum_{k=0}^{m+1} {m \choose k} (-1)^k (x-k)^m + \sum_{k=0}^{m+1} {m \choose k-1} (-1)^k (x-k)^m = \sum_{k=0}^{m} {m \choose k} (-1)^k (x-k)^m + \sum_{k=1}^{m+1} {m \choose k-1} (-1)^k (x-k)^m $
Consider the derivative of the last. It's
$m\sum_{k=0}^{m} {m \choose k} (-1)^k (x-k)^{m-1} + m \sum_{k=1}^{m+1} {m \choose k-1} (-1)^k (x-k)^{m-1} $
Both sums are now under the assumption (we can change variable y = x + 1)
So our derivative is $0$. So it's a constant. Now consider $x = (m+1) / 2$. Our polynomial is clearly 0 here.
southsinger
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