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Knuth (Section 1.2.6) in Example 2 asks

What is the value of $\sum_{k} \binom{n+k}{2k}\binom{2k}{k} \frac{(-1)^k}{k+1}$, if $n$ is a non-negative integer?

Knuth references how to simplify products with Equation (20) $$\binom{r}{m} \binom{m}{k} = \binom{r}{k}\binom{r-k}{m-k}, \text{integer $m$, integer $k$}. \qquad (20)$$ Knuth also references Eq. (7) which is an example of moving in/out of parentheses $$k \binom{r}{k} = r \binom{r-1}{k-1}, \text{integer $k \neq 0$.} \qquad (7)$$

As a first step, Knuth applies Eq. (20) to simplify products. $$\sum_k \binom{n+k}{k}\binom{n}{k} \frac{(-1)^k}{k+1}.$$ The next step is where I don't understand his derivation. Knuth applies Eq. (7) to get $$ \sum_k \binom{n+k}{k} \binom{n+1}{k+1} \frac{(-1)^k}{n+1}. \qquad (27)$$

If you're following along Knuth section (1.2.6) then I added the exact reference which is Eq. (27). I don't see how Knuth derived Eq. (27). I can see how $$\binom{n}{k} k = \binom{n-1}{k-1} n$$ However, it is not clear to me how to apply the approach from Eq. (7) to make $$\binom{n}{k} \frac{(-1)^k}{k+1} = \binom{n+1}{k+1} \frac{(-1)^k}{n+1}$$ What are the steps required to arrive at the solution here?

Tyler
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    So, you ask about the last expression only? Ignore the $(-1)^k$ on both sides as it is the same. Multiply both sides by $(k+1)(n+1)$ leaving us with the proposed identity we wish to check of $\binom{n}{k}(n+1) = \binom{n+1}{k+1}(k+1)$. This can be seen combinatorially in the following way: suppose we have $n+1$ people from which we wish to form a committee of size $k+1$ with one member acting as the committee chair. Do this either by picking the one leader and then the $k$ followers, or by picking the $k+1$ members and one member to be chair. As they count the same scenario they are equal – JMoravitz Mar 03 '22 at 00:39
  • Does this account for $\frac{1}{(n+1)}$ where the last expression needs to result in $\binom{n+1}{k+1} \frac{1}{(n+1)}$ if we're ignoring $(-1)^k$? I see your example but I'm not quite making the connection, thanks! – Tyler Mar 03 '22 at 01:03
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    See https://math.stackexchange.com/questions/3174979/calculate-sum-0-le-k-binomnk2k-binom2kk-frac-1kk1 – Alexander Burstein Mar 03 '22 at 04:32

1 Answers1

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From $(7)$ with $r \mapsto n+1$ and $k \mapsto k+1$, we have $$(k+1)\binom{n+1}{k+1}=(n+1)\binom{n}{k}.$$ Equivalently, $$\frac{1}{n+1}\binom{n+1}{k+1}=\frac{1}{k+1}\binom{n}{k}.$$

RobPratt
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