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Let $$S_r(n)=\sum_{m=0}^n(-1)^m m^r\binom{n}{m}$$ Evaluate $S_n(n)$

For this question, the first part I did is to prove that for integer r, $0<r<n$, $S_r(n)=0$

I did this by induction, given the binomial coefficient $$(1+z)^n=\sum_{m=0}^nz^m \binom{n}{m}$$ and differentiate it $r$ times. The result can be obtained for $S_1(n)=S_2(n)=\dots=S_k(n)=0 \implies S_{k+1}=0, $ given $0<k<n$

I want to show that $S_{n}(n)$ is a linear combination of all $S_k(n)$ for $0<k<n$, but is seems not the case.

How should I proceed?

Martin Sleziak
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Steve
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  • If you just take the binomial expansion of $(1+z)^n$ and differentiate $r$ times, you won't get the sum you're looking for; you get $m(m-1)\cdots(m-r+1)$ instead of $m^r$. You have to differentiate-then-multiply-by-$z$ $r$ times. – Arthur Jan 16 '17 at 12:30

1 Answers1

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I managed to solve it later in the day and here is my solution, using the result that for all $0<r<n$ $$S_r(n)=0$$ So for $S_n(n)$ we differentiate $(1+z)^n$ n times, $$n!=\sum_{m=0}^{n}\binom{n}{m}m(m-1)(m-2)\dots(m-n+1)z^{m-n}$$ Letting $z=-1$, $$n!=\sum_{m=0}^{n}(-1)^m(-1)^n\binom{n}{m}m(m-1)(m-2)\dots(m-n+1)$$ Where the parts with $m$ are just linear combinations of $S_1(n),S_2(n),...S_{n-1}(n)=0$ So $$n!=\sum_{m=0}^{n}(-1)^m\binom{n}{m}[(-1)^nm^n+C_1m^{n-1}+C_2m^{n-2}+...]=(-1)^nS_n(n)$$ which completes the question with $$S_n(n)=n! (-1)^n$$

Steve
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