2

In section 1.9, exercise 19 of the book "Introduction to probability" by Blitzsten and Huang, they prove the following identity with a story proof:

$$\sum\limits_{k=2}^n {k \choose 2}{{n-k+1}\choose 2} = {{n+3}\choose 5}$$

I was looking to prove the same identity with coefficients of polynomials, which is generally a very effective approach for such identities.

My attempt:

The right hand side is the coefficient of $x^5$ in $(1+x)^{n+3}$.

On the left hand side, it's a specific case where the coefficients of $x^2$ are extracted in both expressions. Having a hard time converting this into a polynomial from which coefficients can be extracted.

RobPratt
  • 50,938
Rohit Pandey
  • 7,547
  • 1
    Presumably you meant $\binom{n+2}{5}$, as Marko Riedel's answer verifies. – Semiclassical Mar 09 '25 at 02:51
  • 1
    Also, this problem is a special case (where both binomial coefficients on the left have the same value in the bottom slot) of a more general identity: https://math.stackexchange.com/q/73015/137524, https://math.stackexchange.com/questions/76819/prove-the-identity-sum-limits-s-0-infty-p-s-choose-mqs-choose-q?rq=1. These contains both combinatorial and generating-function proofs. – Semiclassical Mar 09 '25 at 03:12

1 Answers1

2

Write

$$\sum_{k=0}^{n-2} {k+2\choose 2} {n-1-k\choose 2} = \sum_{k=0}^{n-2} {k+2\choose 2} {n-1-k\choose n-3-k} \\ = [z^{n-3}] (1+z)^{n-1} \sum_{k\ge 0} {k+2\choose 2} \frac{z^k}{(1+z)^k} = [z^{n-3}] (1+z)^{n-1} \frac{1}{(1-z/(1+z))^3} \\ = [z^{n-3}] (1+z)^{n-1} (1+z)^3 = [z^{n-3}] (1+z)^{n+2} = {n+2\choose n-3} = {n+2\choose 5}.$$

Here we have on the second line that $z^k/(1+z)^k$ vanishes under the coefficient extractor $[z^{n-3}]$ as soon as $k$ goes past $n-3$ so the extractor enforces the upper range and we may extend $k$ to infinity. This is because $z^k/(1+z)^k = z^k + \cdots$

Commentary, per request. At the second line we start with

$${n-1-k\choose n-3-k} = [z^{n-3-k}] (1+z)^{n-1-k} = [z^{n-3-k}] (1+z)^{n-1} \frac{1}{(1+z)^k} \\ = [z^{n-3}] (1+z)^{n-1} \frac{z^k}{(1+z)^k}.$$

Then we pull the terms that do not depend on $k$ out of the sum. Observe the general rule $[z^{n-k}] f(z) = [z^n] z^k f(z).$

Marko Riedel
  • 64,728
  • Thanks. But I'm having a lot of trouble with the beginning of the second line in your equations. The one where you express the summation as a coefficient of $[z^{n-3}]$. If you have time, could you please elaborate that a bit? – Rohit Pandey Mar 09 '25 at 06:27