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Is there a nice or simple form for a sum of the following form? $$ 1 + \sum_{i=1}^k \binom{n-1+i}{i} - \binom{n-1+i}{i-1}$$

Motivation: Due to a computation in the formalism of Schubert calculus the above sum with $k = \lceil n/2 \rceil -1$ is equal to the number of lines intersecting $2n-4$ general subspaces $H_j\subseteq \mathbb{P}^n$ of dimension $n-2$.

A. Van Werde
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2 Answers2

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Hint

The following method does not use telescopic series but it makes use of elementary binomial theorem along with some geometric progression to arrive at the answer.

$$\begin{aligned}S&=\sum_{i=1}^{k}{n-1+i\choose n-1}-\sum_{i=1}^{k}{n-1+i\choose n}\\&=\left(\text{coeff. of } x^{n-1} \text{ in } \sum_{i=1}^{k}(1+x)^{n-1+i}\right)-\left(\text{coeff. of } x^{n} \text{ in } \sum_{i=1}^{k}(1+x)^{n-1+i}\right)\end{aligned}$$

Paras Khosla
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  • Is there some reason why the sum on the right-hand-side of your first line is equal to mine? Although they look similar at first sight the 'denominator' of the binomial is constant in yours while it varies in mine? – A. Van Werde Dec 20 '19 at 19:31
  • @A.VanWerde I have used the property of binomial coefficients $${n\choose r}={n\choose n-r}$$. – Paras Khosla Dec 20 '19 at 20:04
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No need for a power tool when a manual tool does the trick.

Hockey-stick identity:

$$ \sum_{i=0}^k \binom{n+i}{i} = \binom{n+k+1}{k} $$

Applying to our expression:

$$ 1 + \sum_{i=1}^k \binom{n-1+i}{i} - \binom{n-1+i}{i-1} \\ 1 + \sum_{i=1}^k \binom{n-1+i}{i} - \sum_{i=1}^k \binom{n-1+i}{i-1} \\ 1 - \binom{n-1}{0} + \sum_{i=0}^k \binom{n-1+i}{i} - \sum_{i=0}^k \binom{n+i}{i} \\ \sum_{i=0}^k \binom{n-1+i}{i} - \sum_{i=0}^k \binom{n+i}{i} \\ \binom{n+k}{k} - \binom{n+k+1}{k} \\ \binom{n+k}{k} - \left(\binom{n+k}{k-1}+\binom{n+k}{k}\right) \\ -\binom{n+k}{k-1} $$

zhuli
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  • Do you have a reference to the particular form of the Hockey-stick identity you are using? I can only find it with a constant 'denominator' for the binomials involved. – A. Van Werde Dec 20 '19 at 19:32
  • Apply the symmetric identity for each term: $\binom{n+i}{i} = \binom{n+i}{n}$ – zhuli Dec 20 '19 at 19:57
  • Is anyone going to explain why it's called the Hockey-stick identity?? – Mr Pie Feb 22 '20 at 12:47
  • Its name is derived from the shape the binomial coefficients form when you plot them on Pascal's triangle. https://math.stackexchange.com/questions/3383141/notation-for-nested-sums-of-arbitrary-degree/3383165#3383165 – zhuli Feb 23 '20 at 16:52