Reminiscing about a cute calculus trick

The year was probably 2008. I was a poorly dressed adolescent, with a burgeoning patchy mustache so awful that I still struggle with the shame of having had no desire to shave it off. (No photo posted intentionally, thank you very much.)

I was attending a meeting of the Nassau County All-Star Math Team, which congregated some of the best high school students in the area where I grew up. As practice for upcoming math competitions (NYSML and ARML), we were given problems to work on (we had 10 minutes for every pair of problems), after which students would present their solutions. One such problem really sticks out in my mind, and I was reminded of it while teaching calculus today. I don't remember the exact problem, so here's a version suitable for 2022 (in a manner typical of those competitions). Take a few minutes to try to solve the problem, if you like!

Problem: Let $p(x)$ be the polynomial satisfying $$x^{2022}-5x^{1011}+4 = (x-1) \cdot p(x).$$ Find the sum of the coefficients of $p(x)$.

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After 10 minutes, it was time to present solutions. One student came up and presented the following completely correct method, which is certainly the intended one:

Solution 1: Let us divide the polynomials to compute $p(x)$ explicitly. To see how to do this, we suppose that $p(x) = \sum_{n=0}^{2021}a_n x^n$, and try to compute the coefficients $a_n$ inductively. Notice that $$(x-1) \cdot p(x) = \sum_{n=0}^{2021}a_nx^{n+1} - \sum_{n=0}^{2021}a_nx^n = a_{2021}x^{2022} + \sum_{n=1}^{2021}(a_{n-1}-a_n)x^n - a_0.$$ Hence, we see:

• $a_{2021} = 1$
• $a_{0} = -4$
• $a_{n-1} - a_n = 0$ for any $1 \leq n \leq 2021$ and $n \neq 1011$
• $a_{1010}-a_{1011} = -5$

We can now use this inductively. Notice that most of the time, $a_n = a_{n-1}$, with only the case of $n = 1011$ being special. We find

$$p(x) = x^{2021}+x^{2020} + \cdots + x^{1011} - 4x^{1010} - 4x^{1009} - \cdots - 4x - 4.$$

Therefore, the sum of the coefficients is $(1+1+\cdots + 1) + (- 4 - 4 - \cdots - 4)$ where each of the grouped sums has 1011 terms. Hence, we obtain $$1011 - 4(1011) = \boxed{-3033}.$$

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The coaches then asked if anybody else had any other solution. In fact, I did. I recall walking up to the board with a smirk on my face and an unbridled giddiness bubbling from within.

Solution 2: The sum of the coefficients is

$$p(1) = \lim_{x \rightarrow 1} p(x) = \lim_{x \rightarrow 1}\frac{x^{2022}-5x^{1011}+4 }{x-1} \lim_{x \rightarrow 1} \frac{2022x^{2021} - 5055x^{1010}}{1} = 2022-5055 = \boxed{-3033}$$

by L'Hôpital's rule.

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I finished writing my solution and walked back to my desk to the sound of stunned silence, as though I had violated the sanctity of the high school math competition with such a simple yet unintended method. It looked back at the coaches, whose amusement at my solution made its way from facial expression to audible chuckle after a few seconds. The experience remains my favorite moment from my high school math competition days.