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I haven't seen any similar combinatorial identities of this form in textbooks. I had the left hand side in the first place and the right hand side was given by Mathematica. I'm not sure how Mathematica was able to solve it but I do expect that this has a simple combinatorial proof. I tried to look for one but I was stuck.

Any help?

PS: I am a high school student self-studying combinatorics, so combinatorial proofs or generating-function-based proofs are the best for me. If things like hypergeometric functions or gamma functions cannot be avoided I am glad to accept them as well:) Just means that I need to go deeper. At least it's not some Mathematica magic:)

RobPratt
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Chengyuan Ma
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    Here is the OEIS page for this sequence: https://oeis.org/search?q=2%2C12%2C60%2C280%2C1260%2C5544

    In particular the generating function is $2z(1-4z)^{-3/2}$.

     – Math1000 Jan 14 '20 at 04:25
  • @Math1000 Yes indeed. The generating function for LHS is $\frac{2x}{(2x-1)^2(1-x)^n}$ (fixed $n$). I was stuck when trying to prove that the coefficients of $x^n$ in the expansion of both functions are equal. – Chengyuan Ma Jan 14 '20 at 05:01
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    How do you mean, the generating function for fixed $n$? There's no other free index on the left-hand side – how do you get a generating function if you fix $n$? – joriki Jan 14 '20 at 05:03
  • @joriki Ah I see. Sorry for the mistake. I wrongly fixed $n$ in the binomial... I learned generating function just yesterday for this problem so my understanding is really skin deep =_=. So what should I do to LHS then? – Chengyuan Ma Jan 14 '20 at 05:10
  • This is a special case of this question. – D S Mar 01 '24 at 18:25

3 Answers3

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We seek to show that

$$\sum_{k=1}^n {2n-k-1\choose n-1} k 2^k = n {2n\choose n}.$$

The LHS is

$$\sum_{k=1}^n {2n-k-1\choose n-k} k 2^k \\ = [z^n] (1+z)^{2n-1} \sum_{k=1}^n k 2^k z^k (1+z)^{-k}.$$

Now we may extend $k$ to infinity because of the coefficient extractor in front:

$$[z^n] (1+z)^{2n-1} \sum_{k\ge 1} k 2^k z^k (1+z)^{-k} \\ = [z^n] (1+z)^{2n-1} \frac{2z/(1+z)}{(1-2z/(1+z))^2} \\ = [z^n] (1+z)^{2n} \frac{2z}{(1-z)^2} = 2\sum_{k=0}^n {2n\choose k} (n-k) \\ = 2n \sum_{k=0}^n {2n\choose k} - 2\sum_{k=1}^n {2n\choose k} k = 2n \left(\frac{1}{2} 2^{2n}+\frac{1}{2} {2n\choose n}\right) - 4n\sum_{k=1}^n {2n-1\choose k-1} \\ = n 2^{2n} + n {2n\choose n} - 4n\sum_{k=0}^{n-1} {2n-1\choose k} = n 2^{2n} + n {2n\choose n} - 4n \frac{1}{2} 2^{2n-1} \\ = n {2n\choose n}.$$

This is the claim.

Marko Riedel
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We prose to solve this sum by making use of two results $$\sum_{p=m}^{2m} 2^{-p} {p \choose m}=1~~~~(1)$$ see How to prove that $\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$.

and

see Help in summing: $\sum_{k=m} ^{2m} ~k\cdot ~2^{-k} {k \choose m}$ $$\sum_{p=m}^{2m} p ~2^{-p} {p \choose m}=(2m+1)-2^{-2m-1}(m+1) {2m+2 \choose m+1}~~~~(2)$$

$$S=\sum_{k=1}^{n} k 2 ^k {2n-k-1\choose n-1}= 2^{2m+1} \sum_{p=m}^{2m}[2m+1-p]~2^{-p}~ {p \choose m},~ m=n-1, p=2n-k-1.~~~~~(3)$$ Next by using ((1) and (2) in (3) we get $$S=(m+1) {2m+2 \choose m+1}=n {2n \choose n}.$$

Z Ahmed
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Here’s a combinatorial proof:

The left-hand side sums the position $k$ (starting at $0$ from the left) of the $n$-th element (starting at $1$ from the right) of a subset over all subsets with at least $n$ elements of a set of $2n$ elements: If the $n$-th element is at position $k$, there are $k$ independent binary choices whether to include the $k$ elements to its left in the subset, and exactly $n-1$ of the $2n-k-1$ elements to its right must be included in the subset (since it’s the $n$-th element from the right), for a count of $2^k\binom{2n-k-1}{n-1}$.

We can compute the same sum in a different way: Instead of selecting $j$ out of $2n$ elements in a line, consider selecting $j+1$ out of $2n+1$ elements in a circle, and then selecting one of the $j+1$ selected elements as the point at which to break the circle into a line. In this manner we generate each linear subset of $j$ elements $2n+1$ times. Assign indices $1$ through $2n+1$ to the elements in the circle.

We now sum the positions of the $n$-th element for the $j+1$ linear selections arising from the same circular selection. If we first measure these positions starting from the right, then we need to sum the differences of the circular indices of each of the $j+1$ pairs of selected elements that are $n$ selected elements apart along the circle. The circular index of each selected element appears once as the minuend and once as the substrahend, so these contributions cancel, except that in exactly $n$ cases the difference of the indices is negative and we have to add $2n+1$ to get the actual distance along the circle; thus the sum is $(2n+1)n$. Since we're generating each linear subset $2n+1$ times, we should include a contribution $n$ for every circular selection.

But we measured the positions starting from the right, whereas we actually want the sum of the positions measured starting from the left, so we need to subtract the sum we obtained from $2n$ times the number of linear subsets. Thus, the sum has a contribution $2n$ from every linear subset minus a contribution $n$ from every circular selection:

$$ 2n\sum_{j=n}^{2n}\binom{2n}j-n\sum_{j=n}^{2n}\binom{2n+1}{j+1}\;. $$

In the second sum, we count exactly half the subsets of a set of $2n+1$ elements, so this sum is $\frac122^{2n+1}$. In the first sum, we count half the subsets of a set of $2n$ elements, except the subsets with $n$ elements are counted fully, not just half, so this sum is $\frac122^{2n}+\frac12\binom{2n}n$. Thus our sum is

$$ 2n\left(\frac122^{2n}+\frac12\binom{2n}n\right)-n\cdot\frac122^{2n+1}=n\binom{2n}n\;. $$

joriki
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