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I was reading the recent question on the German tank problem, and had trouble with one of the derivations in this section.

$$\sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom N k} = \frac{k(N+1)}{k+1}$$

I understand where the left-hand side comes from, in the context of the problem. I'm just having trouble arriving at the right-hand side.

\begin{align*} \sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom N k} &= \sum_{m=k}^N m \frac{(m-1)!}{(k-1)!(m-k)!} \frac{k! (N-k)!}{N!}\\ &= k \sum_{m=k}^N \frac{m!}{(m-k)!} \frac{(N-k)!}{N!} \end{align*}

This looks promising, since I have found the $k$ that appears in the right-hand side, but I don't know what to do with the sum.

Community
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angryavian
  • 93,534

2 Answers2

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There is a binomial identity that $\displaystyle \sum_{m=k}^N \binom{m}{k} = \binom{N+1}{k+1}$,

You can see this by induction on $N$, if $N = k$, $\binom{k}{k} = \binom{k+1}{k+1}$, and if it is true that $$\displaystyle \sum_{m=k}^N \binom{m}{k} = \binom{N+1}{k+1}$$ then $$\displaystyle \sum_{m=k}^{N+1} \binom{m}{k} = \binom{N+1}{k+1} + \binom{N+1}{k} = \binom{N+2}{k+1}$$.

Now your sum was $$k \displaystyle \sum_{m=k}^N \frac{m!}{(m-k)!} \frac{(N-k)!}{N!} = \frac{k!k(N-k)!}{N!} \displaystyle \sum_{m=k}^N \binom{m}{k} = \frac{k(N-k)!k!}{N!} \binom{N+1}{k+1} = \frac{k(N+1)}{k+1}$$

lstbl
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Michael Biro
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$$\sum_{m=k}^N m\frac{{m-1 \choose k-1}}{{N \choose k}}=\frac{1}{{N\choose k}}\sum_{m=k}^N m{m-1 \choose k-1}=\frac{1}{{N\choose k}}\sum_{m=k}^{N}k{m \choose k}$$ The last step uses an identity.

Now this equals $$\frac{k}{{N\choose k}}\sum_{m=k}^{N} {m \choose k}=\frac{k}{{N\choose k}}\sum_{m=0}^{N} {m \choose k} =\frac{k}{{N\choose k}}{N+1\choose k+1}$$

vadim123
  • 83,937
  • Looks like I am unfamiliar with a lot of identities on that page... thanks for the answer and the resource! – angryavian Jul 30 '13 at 22:22