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We take: $$(1+x+x^2)^n=a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_{2n}x^{2n}$$ and we need to find the values of the expressions: $$i)a_1+a_4+a_7+a_{10}+\cdots$$ $$ii)a_0-a_2+a_4-a_6+\cdots$$

I have solved similar expressions for eg. $$1) a_0+a_2+a_4+\cdots$$ $$2) a_1+a_3+a_5+\cdots$$ $$3) a_0+a_3+a_6+\cdots$$ by using simple substitutions like $x=1, x=-1, x=\omega, x=\omega^2$ etc. but in these two expressions I'm completely stumped as I've tried using combinations of previous substitutions to create the kinds of expressions in part $i)$ and $ii)$ but was unsuccessful.

Any help would be appreciated.

Jean Marie
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infinite-blank-
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1 Answers1

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Let $f(x) = (1 + x + x^2)^n$, and let $g(x) = f(x)/x$.

  • For 1, consider $g(1) + g(\omega) + g(\omega^2)$, where $\omega = e^{2 \pi i/3}$.
  • For 2, consider $f(i) + f(-i)$.
Ben Grossmann
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  • Thank you for the solution. It might seem like a stupid question but is there any particular way to think about such problems just by looking at the terms of the sequence of coefficients. Like how you assumed g(x)/x seemed like great intuition to me or is there any particular reasoning that goes behind this? – infinite-blank- Aug 04 '20 at 17:26
  • @infinite-blank- You already found the answer for $a_0 + a_3 + a_6 + \cdots$. The same trick doesn't work for $a_1 + a_4 + a_7 + \cdots$ because the coefficients are "shifted over" from where they would need to be. Now, it's easy to see that we can shift the coefficients wherever we'd like by multiplying by a power of $x$, so I shifted the coefficients over so that the first trick would do what we want. – Ben Grossmann Aug 04 '20 at 17:31
  • @infinite-blank- you might find it helpful to consider your "similar problem" 2) following the same line of reasoning. – Ben Grossmann Aug 04 '20 at 17:34
  • As a slight variant, for (i) you could also consider $f(1) + \omega^2 f(\omega) + \omega f(\omega^2)$. – Daniel Schepler Aug 04 '20 at 18:18