Jazmyn Pugh

2022-09-22

How find this $\sum _{k=0}^{2n}\left(-1{\right)}^{k}{a}_{k}^{2}$

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Klecanlh

Expert

The form of the polynomial $p\left(x\right)=1-\sqrt{2}x+{x}^{2}$ appeals to use its conjugate polynomial $q\left(x\right)=1-\sqrt{2}x+{x}^{2}$. Such the usage can be a evaluation of roots and applying Vieta’s formulas, or dealing with a recurrent sequence. But to chose the relevant way to solve the problem we should begin from the end (as recommended George Polya). So, we can see that the asked sum $S=\sum _{k=0}^{2n}\left(-1{\right)}^{k}{a}_{k}^{2}$ is equal to the coefficient at ${x}^{0}$ of the “polynomial” ${p}^{n}\left(x\right){p}^{n}\left(-1/x\right)$. It rests to find it. We have
${p}^{n}\left(x\right){p}^{n}\left(-1/x\right)=\left(p\left(x\right)p\left(-1/x\right){\right)}^{n}=$
$\left(\left(1-\sqrt{2}x+{x}^{2}\right)\left(1/{x}^{2}+\sqrt{2}/x+1\right){\right)}^{n}=$
$\left(\left(1/{x}^{2}\right)\left(1-\sqrt{2}x+{x}^{2}\right)\left(1+\sqrt{2}x+{x}^{2}\right){\right)}^{n}=$
$\left(\left(1/{x}^{2}\right)\left(\left(1+{x}^{2}{\right)}^{2}-\left(\sqrt{2}x{\right)}^{2}\right){\right)}^{n}=\left({x}^{2}+1/{x}^{2}{\right)}^{n}.$
So we see that S=0provided n is odd and $S=\left(\genfrac{}{}{0}{}{n}{n/2}\right)$ provided n is even.

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