Which of the expressions $$(1+x^2-x^3)^{100} \textrm{or}\:\: (1-x^2+x^3)^{100}$$ has the larger coefficient of $x^{20}$ after expending abd and collecting terms.

I can easily do this question via multinomial expansion, but here's the twist. In the question it was written to use the hint given to solve the problem. The hint was replace $-x$ with $x$ to gain extra information.

I replaced, searched for symmetric properties, odd-even properties but couldn't found any. I'm looking for a solution incorporating the above given hint. Also, if there is any method as easy as ABC to do this, I will also appreciate that.


1 Answer 1


The map which replaces $x$ with $-x$ doesn't change coefficients of $x^{2n}$ for integers $n$, and so the expressions have the same $x^{20}$ coefficients as $$(1+x^2+x^3)^{100}\text{ and }(1-x^2-x^3)^{100},$$ respectively. The difference between these two expressions is $$(1+x^2+x^3)^{100}-(1-x^2-x^3)^{100}=\sum_{n=0}^{100}\binom{100}n(1-(-1)^n)(x^2+x^3)^n,$$ which when fully expanded has no term with negative coefficient, and some terms with positive $x^{20}$ coefficient. So, the first polynomial has a greater $x^{20}$ coefficient than the second.

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    $\begingroup$ @TymaGaidash Ah good catch -- I've fixed the upper bound on the sum. $\endgroup$ Sep 30, 2022 at 18:44
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    $\begingroup$ I would just note $(1+x^2+x^3)^{100}$ has positive terms which just add up, while $(1-x^2+x^3)^{100}$ would have same terms, but of both signs hence they wouldn't sum to exceed the first. This is a more specific demonstration though +1. $\endgroup$
    – Macavity
    Sep 30, 2022 at 18:53

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